CBSE Class 12 Chemistry Solutions Assignment Set 04

Question. What mass of NaCl (molar mass = 58.5 g mol^–1) must be dissolved in 65 g of water to lower the freezing point by 7.5°C? The freezing point depression constant, Kf , for water is 1.86 K kg mol^–1. Assume van’t Hoff factor for NaCl is 1.87.
Answer: Given data: \( M_2 = 58.5 \text{ g mol}^{-1} \), \( w_1 = 65 \text{ g} \), \( \Delta T_f = 7.5^\circ\text{C} \), \( K_f = 1.86 \text{ K kg mol}^{-1} \), and \( i = 1.87 \).
Placing these values into the expression:
\( \Delta T_f = i K_f m \)
\( \Delta T_f = i K_f \frac{w_2 \times 1000}{M_2 \times w_1} \)
\( 7.5 = 1.87 \times 1.86 \times \frac{w_2 \times 1000}{58.5 \times 65} \)
\( w_2 = \frac{7.5 \times 58.5 \times 65}{1.87 \times 1.86 \times 1000} \)
\( w_2 = \frac{28518.75}{3478.2} \)
\( \implies w_2 = 8.199 \text{ g} \)
Thus, the total mass of NaCl that needs to be dissolved is \( 8.199 \text{ g} \).
In simple words: To find how much salt is needed to lower water's freezing point, we use the freezing point depression formula and include the van't Hoff factor because salt splits into ions when dissolved.

Exam Tip: Be careful with calculations involving decimals; double-check the numerator and denominator multiplications separately before dividing.

Question. What mass of ethylene glycol (molar mass = 62.0 g mol^–1) must be added to 5.50 kg of water to lower the freezing point of water from 0° C to – 10.0° C? (Kf for water = 1.86 K kg mol–1)?
Answer: Following the main formula for freezing point depression:
\( M_2 = \frac{1000 \times K_f \times w_2}{w_1 \times \Delta T_f} \)
Where \( M_2 = 62 \text{ g mol}^{-1} \), \( w_1 = 5.50 \text{ kg} = 5500 \text{ g} \), \( K_f = 1.86 \text{ K kg mol}^{-1} \), and \( \Delta T_f = 0 - (-10.0) = 10 \text{ K} \).
Substituting these quantities into our equation, we get:
\( 62 = \frac{1000 \times 1.86 \times w_2}{5500 \times 10} \)
\( w_2 = \frac{62 \times 5500 \times 10}{1000 \times 1.86} \)
\( w_2 = \frac{3410}{1.86} \)
\( \implies w_2 = 1833 \text{ g} = 1.833 \text{ kg} \)
Therefore, the required mass of ethylene glycol to be added is \( 1.833 \text{ kg} \).
In simple words: We find the mass of antifreeze needed by rearranging the freezing point depression formula, making sure to convert the water's mass into grams first.

Exam Tip: Always make sure that the units for the mass of the solvent (\( w_1 \)) match the constant \( K_f \), which is expressed in kilograms of solvent.

Question. 15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at – 0.34° C. What is the molar mass of the substance? (Kf for water = 1.86 K kg mol^–1)
Answer: Given information: \( w_2 = 15 \text{ g} \), \( w_1 = 450 \text{ g} \), \( \Delta T_f = 0 - (-0.34) = 0.34 \text{ K} \), \( K_f = 1.86 \text{ K kg mol}^{-1} \), and \( M_2 = ? \).
Placing these parameters into the freezing depression relation:
\( M_2 = \frac{1000 \times K_f \times w_2}{w_1 \times \Delta T_f} \)
\( M_2 = \frac{1000 \times 1.86 \times 15}{450 \times 0.34} \)
\( M_2 = \frac{27900}{153} \)
\( \implies M_2 = 182.53 \text{ g mol}^{-1} \)
So, the molecular weight of the solute is \( 182.53 \text{ g mol}^{-1} \).
In simple words: By measuring how much the freezing point of water drops when we add a known weight of solute, we can calculate its molecular mass.

Exam Tip: Keep your final units for molar mass as \(\text{g mol}^{-1}\) rather than just grams to show a complete understanding of molecular calculations.

Question. What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.5°C? The freezing point depression constant (Kf ) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g)
Answer: Given values: \( M_2 = 58.5 \text{ g mol}^{-1} \), \( w_1 = 65.0 \text{ g} \), \( \Delta T_f = 7.5^\circ\text{C} \), \( K_f = 1.86 \text{ K kg mol}^{-1} \), and \( i = 1.87 \).
By utilizing the depression of freezing point relation:
\( \Delta T_f = i K_f \frac{w_2 \times 1000}{M_2 \times w_1} \)
Substituting these variables yields:
\( 7.5 = 1.87 \times 1.86 \times \frac{w_2 \times 1000}{58.5 \times 65.0} \)
\( w_2 = \frac{7.5 \times 58.5 \times 65.0}{1.87 \times 1.86 \times 1000} \)
\( w_2 = \frac{28518.75}{3478.2} \)
\( \implies w_2 = 8.199 \text{ g} \)
Hence, the weight of salt required to be dissolved is \( 8.199 \text{ g} \).
In simple words: Using the depression of freezing point equation with the van't Hoff factor for table salt, we determine the necessary mass of salt to dissolve.

Exam Tip: Since this is identical to Question 42, ensuring your calculation steps are neat and consistent is key to scoring maximum marks.

Question. Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K. (Kf for water = 1.86 K kg mol^–1)
Answer: Because one mole of KCl yields 2 moles of active ions (\( \text{K}^+ \) and \( \text{Cl}^- \)), the factor \( i \) equals 2. Here, \( \Delta T_f = 2\text{ K} \) and \( K_f = 1.86 \text{ K kg mol}^{-1} \).
Employing the relation:
\( \Delta T_f = i K_f m \)
\( m = \frac{\Delta T_f}{i K_f} = \frac{2}{2 \times 1.86} = \frac{1}{1.86} = 0.54 \text{ mol kg}^{-1} \)
We should mix 0.54 moles of KCl into 1 kg of solvent.
The molecular weight of KCl = 39 + 35.5 = 74.5 g mol\(^{-1}\).
The total mass of KCl needed:
\( \text{Amount of KCl} = 0.54 \times 74.5 \text{ g} = 40.05 \text{ g} \).
In simple words: Since potassium chloride breaks down into two ions in water, we use the freezing point depression formula with a van't Hoff factor of 2 to find the exact mass needed.

Exam Tip: Always calculate the molar mass of ionic salts carefully by adding the correct atomic masses of potassium (39) and chlorine (35.5).

Question. A solution of glycerol (C₃H₈O₃) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42 °C while pure water boils at 100 °C. What mass of glycerol was dissolved to make the solution?
Answer: The change in boiling point is:
\( \Delta T_b = (100.42 - 100)^\circ\text{C} = 0.42^\circ\text{C} = 0.42 \text{ K} \)
Using the elevation formula:
\( \Delta T_b = K_b m \)
\( 0.42 = 0.512 \times \frac{w_2}{92} \times \frac{1000}{500} \)
Here, the molar mass of glycerol (\( \text{C}_3\text{H}_8\text{O}_3 \)) is \( 92 \text{ g mol}^{-1} \), and the solvent weight is \( 500 \text{ g} \).
Rearranging the equation to solve for \( w_2 \):
\( w_2 = \frac{0.42 \times 92 \times 500}{0.512 \times 1000} \)
\( w_2 = \frac{4.83}{0.128} \)
\( \implies w_2 = 37.7 \text{ g} \)
Thus, the total mass of glycerol dissolved in the solution is \( 37.7 \text{ g} \).
In simple words: Adding glycerol raises the boiling point of water. We use the elevation of boiling point formula to work backwards and find the mass of glycerol that was dissolved.

Exam Tip: Keep the value of \( K_b \) for water as 0.512 K kg mol\(^{-1}\) in mind, as it is a standard constant and may sometimes not be explicitly provided in the question.

Question. 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at – 0.34 °C. What is the molar mass of this material? (Kf for water = 1.86 K kg mol^–1)
Answer: By employing the freezing point depression formula:
\( \Delta T_f = K_f \times m \)
\( T_f^\circ - T_f = K_f \times \frac{w_2 \times 1000}{M \times w_1} \)
Substituting our known values yields:
\( [0 - (-0.34)] \text{ K} = 1.86 \text{ K kg mol}^{-1} \times \frac{15\text{ g} \times 1000}{M \times 450\text{ g}} \)
\( M = 1.86 \times \frac{15 \times 1000}{0.34 \times 450} \)
\( M = \frac{27900}{153} \)
\( \implies M = 182 \text{ g mol}^{-1} \)
Hence, the molecular mass of the substance is \( 182 \text{ g mol}^{-1} \).
In simple words: The change in freezing point lets us calculate the molality of the solution, which then tells us the molecular mass of the unknown solute.

Exam Tip: Be precise with units and scientific notation during fractions, as rounding off mid-calculation can slightly alter the final integer value.

Question. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is added to this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate (i) the molecular mass of solute and (ii) vapour pressure of water at 298 K.
Answer: If we assume a highly dilute system, Raoult's law gives:
\( \frac{P^\circ - P}{P^\circ} = \frac{w_B \times M_A}{M_B \times w_A} \)

Case (i):
For the initial scenario, the weight of the solute \( w_B = 30 \text{ g} \), the mass of the solvent (water) \( w_A = 90 \text{ g} \), and the molecular mass of water \( M_A = 18 \text{ g mol}^{-1} \). The vapour pressure \( P = 2.8 \text{ kPa} \).
\( \frac{P^\circ - 2.8}{P^\circ} = \frac{30 \times 18}{M_B \times 90} \)
\( \frac{P^\circ - 2.8}{P^\circ} = \frac{6}{M_B} \) —(1)

Case (ii):
After adding an additional 18 g of water, the total solvent weight becomes \( w_A = 90 + 18 = 108 \text{ g} \), and the new vapour pressure \( P = 2.9 \text{ kPa} \).
\( \frac{P^\circ - 2.9}{P^\circ} = \frac{30 \times 18}{M_B \times 108} \)
\( \frac{P^\circ - 2.9}{P^\circ} = \frac{5}{M_B} \) —(2)

When we divide the first expression (1) by the second (2), we obtain:
\( \frac{P^\circ - 2.8}{P^\circ - 2.9} = \frac{6}{5} \)
\( 5P^\circ - 14.0 = 6P^\circ - 17.4 \)
\( \implies P^\circ = 3.4 \text{ kPa} \)

By inserting this result back into our first equation (1), we find:
\( \frac{3.4 - 2.8}{3.4} = \frac{6}{M_B} \)
\( \frac{0.6}{3.4} = \frac{6}{M_B} \)
\( \implies M_B = \frac{6 \times 3.4}{0.6} = 34 \text{ g mol}^{-1} \trim \)

Therefore, we have:
(i) The molecular mass of the solute is \( 34 \text{ g mol}^{-1} \).
(ii) The vapour pressure of pure water at 298 K is \( 3.4 \text{ kPa} \).
In simple words: We create two equations based on Raoult's law before and after adding more water. Dividing these equations lets us solve for the vapour pressure of pure water and then the mass of the solute.

Exam Tip: Be extremely careful when setting up the ratios in simultaneous equations of relative lowering of vapour pressure to avoid cross-multiplication errors.

Question. If N2 gas is bubbled through water at 293K, how many millimoles of N2 gas would dissolve in 1 litre of water? Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N2 at 293K is 76.48 k bar.
Answer: According to Henry's law:
\( p = K_H \chi \)
\( \chi = \frac{p}{K_H} = \frac{0.987 \text{ bar}}{76.48 \times 10^3 \text{ bar}} = 1.29 \times 10^{-5} \)
For 1 litre of water, the number of moles of water is:
\( n_{\text{H}_2\text{O}} = \frac{1000 \text{ g}}{18 \text{ g mol}^{-1}} = 55.5 \text{ mol} \)
The mole fraction of nitrogen is expressed as:
\( \chi_{\text{N}_2} = \frac{n_{\text{N}_2}}{n_{\text{N}_2} + n_{\text{H}_2\text{O}}} \)
Since the number of dissolved nitrogen moles is extremely small compared to water, we omit this value from the bottom of the fraction (\( n_{\text{N}_2} \ll 55.5 \)):
\( \chi_{\text{N}_2} \approx \frac{n_{\text{N}_2}}{55.5} \)
\( 1.29 \times 10^{-5} = \frac{n_{\text{N}_2}}{55.5} \)
\( \implies n_{\text{N}_2} = 1.29 \times 10^{-5} \times 55.5 = 7.16 \times 10^{-4} \text{ mol} \)
Converting moles to millimoles:
\( \text{Millimoles of } \text{N}_2 = 7.16 \times 10^{-4} \times 1000 = 0.716 \text{ mmol} \).
In simple words: Henry's law tells us the fraction of nitrogen in water. Since the amount is very small, we can simplify our fraction to find the exact millimoles dissolved in one litre.

Exam Tip: Remember to convert the Henry's law constant from kilobars to bars so that its unit matches the unit of the partial pressure.

Question. The partial pressure of ethane over a saturated solution containing 6.56 × 10-2 g of ethane is 1 bar. If the solution contains 5.0 × 10–2 g of ethane, then what will be the partial pressure of the gas?
Answer: According to Henry’s law, the mass of a dissolved gas is proportional to its partial pressure:
\( m = K_H \times p \)

In the first case:
\( 6.56 \times 10^{-2} \text{ g} = K_H \times 1 \text{ bar} \)
\( \implies K_H = 6.56 \times 10^{-2} \text{ g bar}^{-1} \)

Substituting the constant \( K_H \) into our second equation yields:
\( 5.0 \times 10^{-2} \text{ g} = 6.56 \times 10^{-2} \text{ g bar}^{-1} \times p \)
\( p = \frac{5.0 \times 10^{-2}}{6.56 \times 10^{-2}} \)
\( \implies p = 0.762 \text{ bar} \)
Thus, the partial pressure of the gas is \( 0.762 \text{ bar} \).
In simple words: Since the weight of dissolved gas is directly related to pressure, we can use the ratio of the two weights to calculate the new pressure directly.

Exam Tip: Since \( K_H \) cancels out in direct proportion problems, you can also solve this quickly as \( \frac{m_1}{p_1} = \frac{m_2}{p_2} \).

Question. Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in 2L of water at 25° C, assuming that it is completely dissociated. (R = 0.0821 L atm K^–1 mol^–1, Molar mass of K2SO4 = 174 g mol^–1).
Answer: The relation for osmotic pressure is:
\( \pi = i C R T = i \frac{w}{M V} R T \)
Given: \( w = 2.5 \times 10^{-2} \text{ g} = 0.025 \text{ g} \), \( V = 2 \text{ L} \), \( T = 25^\circ\text{C} = 298 \text{ K} \), and molar mass \( M = 174 \text{ g mol}^{-1} \).
Under the assumption of complete ionization:
\( \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \)
The number of ions produced is 3, which means \( i = 3 \).
Substituting these values into the osmotic pressure formula:
\( \pi = 3 \times \frac{0.025 \text{ g}}{174 \text{ g mol}^{-1} \times 2 \text{ L}} \times 0.0821 \text{ L atm K}^{-1}\text{mol}^{-1} \times 298 \text{ K} \)
\( \pi = \frac{1.835}{348} \)
\( \implies \pi = 5.27 \times 10^{-3} \text{ atm} \).
In simple words: Since potassium sulfate splits into three particles in water, its osmotic pressure is three times higher than if it stayed together. We plug these numbers into our pressure equation.

Exam Tip: Be sure to write the balanced dissociation equation for ionic compounds to correctly determine the van't Hoff factor \( i \).

Question. The partial pressure of ethane over a saturated solution containing 6.56 × 10–2 g of ethane is 1 bar. If the solution were to contain 5.0 × 10–2 g of ethane, then what will be the partial pressure of the gas?
Answer: Using Henry's Law, the mass of dissolved gas is directly related to its partial pressure:
\( m = K_H \times p \)
In the initial condition:
\( 6.56 \times 10^{-2} \text{ g} = K_H \times (1 \text{ bar}) \)
\( \implies K_H = 6.56 \times 10^{-2} \text{ g bar}^{-1} \)

Applying this constant to the second condition:
\( 5.0 \times 10^{-2} \text{ g} = \left(6.56 \times 10^{-2} \text{ g bar}^{-1}\right) \times p \)
\( p = \frac{5.0 \times 10^{-2}}{6.56 \times 10^{-2}} \)
\( \implies p = 0.762 \text{ bar} \)
Therefore, the final partial pressure is determined to be \( 0.762 \text{ bar} \).
In simple words: This question is identical to Question 51. Since the dissolved gas weight depends linearly on pressure, the new pressure is easily computed.

Exam Tip: Stating the proportional relationship clearly before performing calculations is a great way to ensure you gain full partial credits.

Question. Some ethylene glycol, HOCH2CH2OH, is added to your car’s cooling system along with 5 kg of water. If the freezing point of water-glycol solution is –15.0°C, what is the boiling point of the solution? (Kb = 0.52 K kg mol^–1 and Kf = 1.86 K kg mol^–1 for water)
Answer: Given data: \( \Delta T_f = 0 - (-15.0) = 15^\circ\text{C} \) (or \( 15 \text{ K} \)), solvent weight \( w_1 = 5 \text{ kg} = 5000 \text{ g} \), constant \( K_f = 1.86 \text{ K kg mol}^{-1} \), and constant \( K_b = 0.52 \text{ K kg mol}^{-1} \).
Using the formula:
\( \Delta T_f = \frac{1000 \times K_f \times w_2}{w_1 \times M_2} \)
Where the molecular mass of ethylene glycol (\( \text{C}_2\text{H}_6\text{O}_2 \)) is \( 62 \text{ g mol}^{-1} \).
Let us determine the weight of the solute \( w_2 \):
\( w_2 = \frac{\Delta T_f \times w_1 \times M_2}{1000 \times K_f} = \frac{15 \times 5000 \times 62}{1000 \times 1.86} \)
\( \implies w_2 = 2500 \text{ g} \)

Now, we calculate the elevation in boiling point \( \Delta T_b \):
\( \Delta T_b = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1} \)
\( \Delta T_b = 0.52 \times \frac{2500 \times 1000}{62 \times 5000} \)
\( \implies \Delta T_b = 4.19^\circ\text{C} \) (or \( 4.19 \text{ K} \))

Since pure water boils at \( 100^\circ\text{C} \) (or \( 373.15 \text{ K} \)), the boiling point of the solution is:
\( T_b = 100^\circ\text{C} + 4.19^\circ\text{C} = 104.19^\circ\text{C} \) (or \( 377.34 \text{ K} \)).
In simple words: The change in freezing point tells us how much solute is in the water. We then use that amount of solute to find out how much the boiling point of the water will go up.

Exam Tip: Be sure to add the boiling point elevation to the standard boiling point of water (100°C or 373.15 K) to get the final solution boiling point, rather than leaving just the elevation value.

Question. 3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the Van’t Hoff factor and predict the nature of solute (associated or dissociated). (Given : Molar mass of benzoic acid = 122 g mol^–1, Kf for benzene = 4.9 K kg mol^–1)
Answer: The relation for freezing point depression including the van't Hoff factor is:
\( \Delta T_f = i K_f \times \frac{w_b \times 1000}{m_b \times w_a} \)
Substituting our known values:
\( 1.62 = i \times 4.9 \times \frac{3.9 \times 1000}{122 \times 49} \)
Rearranging the equation to solve for \( i \):
\( i = \frac{1.62 \times 122 \times 49}{4.9 \times 3.9 \times 1000} \)
\( i = \frac{9684.36}{19110} \)
\( \implies i = 0.506 \)
With \( i \) less than 1, we can conclude that the solute molecules undergo association (dimerization) in the solution.
In simple words: If the van't Hoff factor is less than one, it means the molecules are clustering together (associating) in the solvent, reducing the effective number of particles.

Exam Tip: Clearly state the rule that \( i < 1 \) implies association and \( i > 1 \) implies dissociation to secure full marks for the qualitative part of the question.

Question. A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute. (Vapour pressure of pure water at 308 K = 32 mm Hg)
Answer: Following Raoult's law for a dilute solution:
\( \frac{P^\circ_{\text{solvent}} - P_{\text{solvent}}}{P^\circ_{\text{solvent}}} = \frac{w_{\text{solute}} \times M_{\text{solvent}}}{M_{\text{solute}} \times w_{\text{solvent}}} \)
Substituting the given values into the formula:
\( \frac{32.00 - 31.84}{32} = \frac{10 \times 18}{M_{\text{solute}} \times 200} \)
\( \frac{0.16}{32} = \frac{9}{10 M_{\text{solute}}} \)
\( M_{\text{solute}} = \frac{32 \times 9}{10 \times 0.16} \)
\( M_{\text{solute}} = \frac{288}{1.6} \)
\( \implies M_{\text{solute}} = 180 \text{ g mol}^{-1} \)
Thus, the molar mass of the non-volatile solute is \( 180 \text{ g mol}^{-1} \).
In simple words: By comparing the drop in vapour pressure when a solute is added, we find the ratio of solute moles to solvent moles and use it to solve for the solute's molecular mass.

Exam Tip: Remember that the molar mass of the solvent (water) is a standard \( 18 \text{ g mol}^{-1} \). Keeping calculations of relative lowering clean avoids fractional mistakes.

Question. 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate (i) the freezing point depression and (ii) the freezing point of the solution (Given : Kf of water = 1.86 K kg mol^–1)
Answer: (i) Given: \( w = 45 \text{ g} \trim \), \( W = 600 \text{ g} \), and \( K_f = 1.86 \text{ K kg mol}^{-1} \).
The molecular mass of ethylene glycol (\( \text{C}_2\text{H}_6\text{O}_2 \)) is:
\( M = 2 \times 12 + 6 \times 1 + 2 \times 16 = 62 \text{ g mol}^{-1} \).
Employing the formula for freezing point depression:
\( \Delta T_f = K_f \times \frac{w \times 1000}{M \times W} \)
\( \Delta T_f = 1.86 \times \frac{45 \times 1000}{62 \times 600} \)
\( \Delta T_f = \frac{837}{372} \)
\( \implies \Delta T_f = 2.25 \text{ K} \)

(ii) The freezing point of the solution is given by:
\( \Delta T_f = T_f^\circ - T_f \)
Where \( T_f^\circ \) of pure water is \( 273.15 \text{ K} \) (or \( 0^\circ\text{C} \)).
\( 2.25 \text{ K} = 273.15 \text{ K} - T_f \)
\( T_f = 273.15 - 2.25 \)
\( \implies T_f = 270.9 \text{ K} \).
In simple words: First we find how much the freezing point drops by using the concentration of antifreeze, and then we subtract this change from the normal freezing point of pure water.

Exam Tip: Be sure to explicitly state both the freezing point depression (\( \Delta T_f \)) and the final freezing point (\( T_f \)) to fully answer both parts of the question.

Question. A 5 percent solution (by mass) of cane-sugar (M.W. 342) is isotonic with 0.877% solution of substance X. Find the molecular weight of X.
Answer: Isotonic solutions share the same osmotic pressure at a given temperature, meaning their molar concentrations are identical (\( C_1 = C_2 \)):
\( \frac{w_{\text{sugar}}}{M_{\text{sugar}} \times V} = \frac{w_X}{M_X \times V} \)
\( \implies \frac{w_{\text{sugar}}}{M_{\text{sugar}}} = \frac{w_X}{M_X} \)
Given: sugar concentration is 5% (5 g in 100 mL) and substance X is 0.877% (0.877 g in 100 mL), and \( M_{\text{sugar}} = 342 \text{ g mol}^{-1} \).
Substituting these values into the ratio:
\( \frac{5}{342} = \frac{0.877}{M_X} \)
\( M_X = \frac{0.877 \times 342}{5} \)
\( M_X = \frac{299.934}{5} \)
\( \implies M_X = 59.98 \approx 60 \text{ g mol}^{-1} \).
Thus, the molecular weight of substance X is approximately \( 60 \text{ g mol}^{-1} \).
In simple words: Solutions that are isotonic have the exact same concentration of dissolved particles. Setting their concentrations equal allows us to solve for the missing molar mass.

Exam Tip: Remember to state the core condition of isotonicity (\( \pi_1 = \pi_2 \) or \( C_1 = C_2 \)) first to demonstrate your theoretical foundation before starting calculations.

Question. Calculate the boiling point of solution when 4 g of MgSO4 (M =120 g mol^–1) was dissolved in 100 g of water, assuming MgSO4 undergoes complete ionization. (Kb for water = 0.52 K kg mol^–1)
Answer: Because magnesium sulfate is an ionic electrolyte, it undergoes complete dissociation:
\( \text{MgSO}_4 \rightarrow \text{Mg}^{2+} + \text{SO}_4^{2-} \)
The total moles of particles produced is 2, hence \( i = 2 \).
Applying the boiling point elevation formula:
\( \Delta T_b = i K_b m = i K_b \frac{w_2 \times 1000}{M_2 \times w_1} \)
Substituting our known parameters:
\( \Delta T_b = 2 \times 0.52 \times \frac{4 \times 1000}{120 \times 100} \)
\( \Delta T_b = \frac{4160}{12000} \)
\( \implies \Delta T_b = 0.346 \text{ K} \)

Since pure water boils at \( 373.15 \text{ K} \), the boiling point of the solution is:
\( T_b = T_b^\circ + \Delta T_b = 373.15 \text{ K} + 0.346 \text{ K} = 373.496 \text{ K} \).
In simple words: Magnesium sulfate splits into two separate ions in water, making the boiling point elevation twice as large. We find the elevation and add it to water's normal boiling point.

Exam Tip: Be careful with ionic compounds like MgSO4; forgetting to multiply by the van't Hoff factor \( i = 2 \) is a very common error.

Question. Calculate the mass of a non-volatile solute (molecular mass 40) which should be dissolved in 114 g octane to reduce the vapour pressure to 80%.
Answer: Let the vapour pressure of pure octane be \( P^\circ = 100 \text{ mm Hg} \). To lower it to 80%, the solution vapour pressure must be \( P = 80 \text{ mm Hg} \).
Using Raoult's law equation:
\( \frac{P^\circ - P}{P^\circ} = \frac{w_B \times M_A}{M_B \times w_A} \)
The solvent is octane (\( \text{C}_8\text{H}_{18} \)) with molecular weight \( M_A = 8 \times 12 + 18 \times 1 = 114 \text{ g mol}^{-1} \), and solvent mass \( w_A = 114 \text{ g} \).
Substituting these values into our relation:
\( \frac{100 - 80}{100} = \frac{w_B \times 114}{40 \times 114} \)
\( \frac{20}{100} = \frac{w_B}{40} \)
\( \implies w_B = \frac{20 \times 40}{100} = 8 \text{ g} \).
Thus, the total mass of the solute required is \( 8 \text{ g} \).
In simple words: To drop the vapour pressure by 20%, we match the ratio of lowering to the mole ratio of solute and solvent, solving for the mass of the solute.

Exam Tip: Recognize that reducing the pressure "to 80%" means the lowering (\( P^\circ - P \)) is exactly 20% of the initial pressure.

Question. An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute? [Vapour pressure of water = 1.013 bar]
Answer: For a 2% by mass aqueous solution, the mass of solute \( w_B = 2 \text{ g} \) and the mass of solvent (water) \( w_A = 98 \text{ g} \). The molecular mass of the solvent is \( M_A = 18 \text{ g mol}^{-1} \).
Using Raoult's law for a dilute solution:
\( \frac{P^\circ - P}{P^\circ} = \frac{w_B \times M_A}{M_B \times w_A} \)
Substituting our known parameters:
\( \frac{1.013 - 1.004}{1.013} = \frac{2 \times 18}{M_B \times 98} \)
\( \frac{0.009}{1.013} = \frac{36}{98 M_B} \)
\( M_B = \frac{36 \times 1.013}{0.009 \times 98} \)
\( M_B = \frac{36.468}{0.882} \)
\( \implies M_B = 41.3 \text{ g mol}^{-1} \).
Thus, the molecular mass of the non-volatile solute is \( 41.3 \text{ g mol}^{-1} \).
In simple words: We find the lowering of water's vapour pressure when a small amount of solute is added, and use Raoult's law to solve for the solute's molecular mass.

Exam Tip: Ensure that your mass of solvent is correct (100 - 2 = 98 g) for percentage-based solutions rather than using the full 100 g.

Question. A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K. Given: (Molar mass of sucrose = 342 g mol^–1) (Molar mass of glucose = 180 g mol^–1)
Answer: For a 10% by mass solution of either solute, the mass of solute \( w = 10 \text{ g} \) and the mass of solvent (water) \( W = 90 \text{ g} \).

For Sucrose:
Molar mass of sucrose (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)) is \( 342 \text{ g mol}^{-1} \).
Molality of sucrose \( m_{\text{sucrose}} = \frac{10 \times 1000}{90 \times 342} = 0.3244 \text{ mol kg}^{-1} \).
Depression in freezing point \( \Delta T_f = T_f^\circ - T_f = 273.15 - 269.15 = 4 \text{ K} \).
Using the relationship \( \Delta T_f = K_f \times m \):
\( K_f = \frac{\Delta T_f}{m} = \frac{4}{0.3244} = 12.33 \text{ K kg mol}^{-1} \).

For Glucose:
Molar mass of glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) is \( 180 \text{ g mol}^{-1} \).
Molality of glucose \( m_{\text{glucose}} = \frac{10 \times 1000}{90 \times 180} = 0.6166 \text{ mol kg}^{-1} \).
Using the value of \( K_f \) calculated above:
\( \Delta T_f = K_f \times m_{\text{glucose}} = 12.33 \times 0.6166 = 7.60 \text{ K} \).

Therefore, the freezing point of the glucose solution is:
\( T_f = T_f^\circ - \Delta T_f = 273.15 \text{ K} - 7.60 \text{ K} = 265.55 \text{ K} \).
In simple words: First we use the freezing point of the sucrose solution to find the constant constant of water, and then we use that constant to calculate the new freezing point for glucose.

Exam Tip: Since both solutions are prepared in water, the freezing point depression constant (\( K_f \)) remains exactly the same for both calculations.

Question. The vapour pressure of pure liquids A and B at 400 K are 450 and 700 mmHg respectively. Find out the composition of liquid mixture if total vapour pressure at this temperature is 600 mmHg.
Answer: Given data: \( P_A^\circ = 450 \text{ mm Hg} \), \( P_B^\circ = 700 \text{ mm Hg} \), and \( P_{\text{total}} = 600 \text{ mm Hg} \).
Applying Raoult's law for the total vapour pressure:
\( P_{\text{total}} = x_A P_A^\circ + x_B P_B^\circ \)
Since \( x_A + x_B = 1 \implies x_B = 1 - x_A \):
\( P_{\text{total}} = x_A P_A^\circ + (1 - x_A) P_B^\circ \)
\( 600 = 450 x_A + 700 (1 - x_A) \)
\( 600 = 700 - 250 x_A \)
\( 250 x_A = 100 \)
\( \implies x_A = 0.40 \)

The liquid phase mole fraction of B is:
\( x_B = 1 - 0.40 = 0.60 \).

Now, let us calculate the partial vapour pressures of both components:
\( P_A = x_A \times P_A^\circ = 0.40 \times 450 = 180 \text{ mm Hg} \)
\( P_B = x_B \times P_B^\circ = 0.60 \times 700 = 420 \text{ mm Hg} \)

Using these, we can find the mole fractions of both components in the vapour phase:
\( y_A = \frac{P_A}{P_{\text{total}}} = \frac{180}{600} = 0.30 \)
\( y_B = \frac{P_B}{P_{\text{total}}} = \frac{420}{600} = 0.70 \).
Therefore, the composition of the mixture is:
- Liquid phase: \( x_A = 0.40 \), \( x_B = 0.60 \)
- Vapour phase: \( y_A = 0.30 \trim \), \( y_B = 0.70 \).
In simple words: Raoult's law helps us find the mole fraction of each liquid in the mixture. We then divide their partial pressures by the total pressure to find their concentrations in the gas phase above the liquid.

Exam Tip: Be sure to calculate the composition in both the liquid phase (\(x\)) and vapour phase (\(y\)) to provide a complete answer.

Question. (a) Define the following terms :
(i) Mole fraction (ii) Van’t Hoff factor
(b) 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of protein? (R = 0.0821 L atm mol^–1 K^–1 and 760 mm Hg = 1 atm)
Answer:
(a) definitions:
(i) Mole fraction: It is defined as the ratio of the number of moles of a single component in a mixture to the total number of moles of all constituents present in the solution. It is commonly denoted by \( x \).
(ii) Van't Hoff factor: It is the ratio of the experimentally observed colligative property of a solution to its calculated or theoretical value. It is represented by \( i \).

(b) Numerical Calculation:
Given: mass of solute \( w_2 = 100 \text{ mg} = 0.1 \text{ g} \), volume of solution \( V = 10.0 \text{ mL} = 0.01 \text{ L} \), temperature \( T = 25^\circ\text{C} = 298 \text{ K} \), and osmotic pressure:
\( \pi = 13.3 \text{ mm Hg} = \frac{13.3}{760} \text{ atm} \approx 0.0175 \text{ atm} \).
Using the relation for osmotic pressure:
\( \pi = \frac{w_2 R T}{M_2 V} \)
\( M_2 = \frac{w_2 R T}{\pi V} \)
\( M_2 = \frac{0.1 \times 0.0821 \times 298}{\left(\frac{13.3}{760}\right) \times 0.01} \)
\( M_2 = \frac{0.1 \times 0.0821 \times 298 \times 760}{13.3 \times 0.01} \)
\( M_2 = \frac{1860.03}{0.133} \)
\( \implies M_2 = 13985 \text{ g mol}^{-1} \).
Hence, the molar mass of the protein is \( 13985 \text{ g mol}^{-1} \).
In simple words: After defining the two terms, we convert mass to grams and pressure to atmospheres to calculate the molecular mass of the protein using the osmotic pressure formula.

Exam Tip: High-molecular-weight polymers like proteins exhibit low osmotic pressures, so keeping unit conversions like torr/mmHg to atm highly accurate is very important.

Question. (a) What is meant by :
(i) Colligative properties (ii) Molality of a solution
(b) What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of 25° C, a total pressure of 1 atmosphere and mole fraction of nitrogen in air of 0.78. [KH for nitrogen = 8.42 × 10–7 M/mm Hg]
Answer:
(a) definitions:
(i) Colligative properties: Those physical properties of solutions that depend solely on the relative number of solute particles present in the solution and are independent of their chemical nature.
(ii) Molality: The total number of moles of solute dissolved per 1000 grams (1 kg) of the solvent. It is denoted as \( m \).

(b) Numerical Calculation:
Given values: Total pressure \( P_{\text{total}} = 1 \text{ atm} = 760 \text{ mm Hg} \), mole fraction of nitrogen in air \( x_{\text{N}_2} = 0.78 \), and Henry's constant \( K_H = 8.42 \times 10^{-7} \text{ M/mm Hg} \).
First, find the partial pressure of nitrogen gas (\( P_{\text{N}_2} \)):
\( P_{\text{N}_2} = x_{\text{N}_2} \times P_{\text{total}} \)
\( P_{\text{N}_2} = 0.78 \times 760 \text{ mm Hg} = 592.8 \text{ mm Hg} \).
Using Henry's law to solve for concentration \( C_{\text{N}_2} \):
\( C_{\text{N}_2} = K_H \times P_{\text{N}_2} \)
\( C_{\text{N}_2} = 8.42 \times 10^{-7} \text{ M/mm Hg} \times 592.8 \text{ mm Hg} \)
\( \implies C_{\text{N}_2} = 4.99 \times 10^{-4} \text{ M} \).
Thus, the concentration of nitrogen in the water is \( 4.99 \times 10^{-4} \text{ mol L}^{-1} \).
In simple words: We find the partial pressure of nitrogen in the air above the glass of water first. Then, multiplying it by the Henry's law constant gives us the dissolved concentration of nitrogen gas.

Exam Tip: Be mindful of the unit of Henry's constant (\(\text{M/mm Hg}\)), which means you must express the pressure in mm Hg rather than atmospheres before multiplying.

Question. (a) Differentiate between molarity and molality for a solution. How does a change in temperature influence their values?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g) (Kf for water = 1.86 K kg mol^–1)
Answer:
(a) Differences and temperature influence:
- Molarity (M): Measures the number of moles of solute in 1 L of solution. It depends on temperature because volume expands or contracts when heated or cooled.
- Molality (m): Measures the number of moles of solute dissolved in 1 kg of solvent. It is independent of temperature because mass does not change when temperature varies.

(b) Numerical Calculation:
Since \( \text{MgBr}_2 \) is an ionic salt, it undergoes complete dissociation in water:
\( \text{MgBr}_2 \rightarrow \text{Mg}^{2+} + 2\text{Br}^- \)
The total moles of ions produced is 3, so \( i = 3 \).
Applying the depression formula:
\( \Delta T_f = i K_f m \)
\( \Delta T_f = i K_f \frac{w_2 \times 1000}{M_2 \times w_1} \)
\( \Delta T_f = 3 \times 1.86 \times \frac{10.50 \times 1000}{184 \times 200} \)
\( \Delta T_f = \frac{58590}{36800} \)
\( \implies \Delta T_f = 1.59^\circ\text{C} \) (or \( 1.59 \text{ K} \))

Therefore, the freezing point of the solution is:
\( T_f = T_f^\circ - \Delta T_f = 0^\circ\text{C} - 1.59^\circ\text{C} = -1.59^\circ\text{C} \) (or \( 271.41 \text{ K} \)).
In simple words: Molarity changes with temperature because volume shifts, but molality stays constant because mass does not. We calculate the freezing point by multiplying the concentration by the van't Hoff factor.

Exam Tip: Be sure to write the complete dissociation equation to show the examiner how you got the van't Hoff factor value of 3.

Question. (a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (Kb for water = 0.512 K kg mol^–1, Molar mass of NaCl = 58.44 g)
Answer:
(a) definitions and explanation:
- Osmosis: The passive movement of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semi-permeable membrane.
- Osmotic pressure: The minimum external pressure that must be applied to the solution to prevent the inward flow of solvent across the semi-permeable membrane.
- Colligative property explanation: Yes, osmotic pressure is a colligative property because its magnitude depends solely on the concentration of solute particles present in the solution, regardless of their chemical identity.

(b) Numerical Calculation:
Since \( \text{NaCl} \) is an electrolyte, it dissociates completely in solution:
\( \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \)
The total moles of particles produced is 2, hence \( i = 2 \).
Using the boiling point elevation formula:
\( \Delta T_b = i K_b m = i K_b \frac{w_2 \times 1000}{M_2 \times w_1} \)
\( T_b - T_b^\circ = 2 \times 0.512 \times \frac{15.00 \times 1000}{58.44 \times 250.0} \)
\( T_b - T_b^\circ = \frac{15360}{14610} \)
\( \implies T_b - T_b^\circ = 1.05 \text{ K} \)

Since pure water boils at \( 373.15 \text{ K} \):
\( T_b = 373.15 \text{ K} + 1.05 \text{ K} = 374.20 \text{ K} \) (or \( 101.05^\circ\text{C} \)).
In simple words: Osmosis is the flow of liquid through a membrane, and osmotic pressure is the force needed to stop it. We find the boiling point of salt water by adding the calculated boiling elevation to pure water's boiling point.

Exam Tip: Remember that adding any non-volatile solute always raises the boiling point and lowers the freezing point of the solvent.

Question. (a) State the following :
(i) Henry’s law about partial pressure of a gas in a mixture.
(ii) Raoult’s law in its general form in reference to solutions.
(b) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is a non-electrolyte, determine its molar mass.
Answer:
(a) Laws:
(i) Henry's law: The solubility of a gas in a liquid solvent at a constant temperature is directly proportional to the partial pressure of that gas in equilibrium with the liquid.
(ii) Raoult's law: For any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction in the liquid mixture.

(b) Numerical Calculation:
Given: solute mass \( w_2 = 8.95 \text{ mg} = 8.95 \times 10^{-3} \text{ g} \trim \), volume \( V = 35.0 \text{ mL} = 0.035 \text{ L} \), temperature \( T = 25^\circ\text{C} = 298 \text{ K} \), and pressure:
\( \pi = 0.335 \text{ torr} = \frac{0.335}{760} \text{ atm} \approx 4.41 \times 10^{-4} \text{ atm} \).
Using the osmotic pressure formula:
\( \pi = \frac{w_2 R T}{M_2 V} \)
\( M_2 = \frac{w_2 R T}{\pi V} \)
\( M_2 = \frac{\left(8.95 \times 10^{-3} \text{ g}\right) \times 0.0821 \text{ L atm K}^{-1}\text{mol}^{-1} \times 298 \text{ K}}{\left(\frac{0.335}{760}\text{ atm}\right) \times 0.035 \text{ L}} \)
\( M_2 = \frac{8.95 \times 10^{-3} \times 0.0821 \times 298 \times 760}{0.335 \times 0.035} \)
\( M_2 = \frac{166.416}{0.011725} \)
\( \implies M_2 = 1.42 \times 10^4 \text{ g mol}^{-1} \).
Therefore, the molecular mass of the gene fragment is \( 1.42 \times 10^4 \text{ g mol}^{-1} \).
In simple words: Henry's and Raoult's laws explain how pressure relates to gas and liquid solubility. We find the molecular mass of the large gene fragment by carefully converting pressure and mass units and using the osmotic pressure equation.

Exam Tip: Be sure to write the formula in terms of \(M_2\) before plugging in values, as this reduces algebraic confusion when handling multiple conversion factors.

Question. (a) Differentiate between molarity and molality in a solution. What is the effect of temperature change on molarity and molality in a solution?
(b) What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g of chloroform form a solution that has a boiling point of 68.04°C. The boiling point of pure chloroform is 61.7°C and the boiling point elevation constant, Kb for chloroform is 3.63°C/m.
Answer:
(a) Differences and temperature influence:
- Molarity (M): Represents the moles of solute dissolved in 1 L of solution. It varies with temperature because liquid volume is temperature-dependent.
- Molality (m): Represents the moles of solute dissolved in 1 kg of solvent. It is independent of temperature because mass does not change as temperature fluctuates.

(b) Numerical Calculation:
Given: mass of solute \( w_2 = 6.21 \text{ g} \), mass of solvent (chloroform) \( w_1 = 24.0 \text{ g} \), boiling point of solution \( T_b = 68.04^\circ\text{C} \), and boiling point of pure solvent \( T_b^\circ = 61.7^\circ\text{C} \).
Calculate the elevation in boiling point \( \Delta T_b \):
\( \Delta T_b = T_b - T_b^\circ = 68.04 - 61.7 = 6.34^\circ\text{C} \) (or \( 6.34 \text{ K} \)).
Given the boiling constant \( K_b = 3.63^\circ\text{C/m} \), we use the boiling elevation equation:
\( M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1} \)
\( M_2 = \frac{3.63 \times 6.21 \times 1000}{6.34 \times 24.0} \)
\( M_2 = \frac{22542.3}{152.16} \)
\( \implies M_2 = 148.14 \text{ g mol}^{-1} \).
Thus, the molar mass of the compound is \( 148.14 \text{ g mol}^{-1} \).
In simple words: Temperature changes molarity because liquids expand, but it doesn't affect molality. We calculate the unknown molecular weight using the formula for elevation of boiling point.

Exam Tip: Chloroform's boiling point constant is quite high compared to water; double check your calculations to make sure the units match perfectly.

Question. (a) Define the following terms :
(i) Mole fraction (ii) Ideal solution
(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at – 0.34°C. What is the molar mass of the material? (Kf for water = 1.86 K kg mol^–1)
Answer:
(a) definitions:
(i) Mole fraction: The ratio of the number of moles of one component to the total number of moles of all the components present in the mixture.
\( x_A = \frac{n_A}{n_A + n_B} \)
(ii) Ideal solution: A solution that strictly obeys Raoult's law at all concentrations and across all temperatures.

(b) Numerical Calculation:
Given: solute mass \( w_2 = 15.0 \text{ g} \), solvent mass (water) \( w_1 = 450 \text{ g} \), constant \( K_f = 1.86 \text{ K kg mol}^{-1} \), and freezing point depression:
\( \Delta T_f = 0 - (-0.34) = 0.34 \text{ K} \).
Using the freezing depression formula:
\( \Delta T_f = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1} \)
\( M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} \)
\( M_2 = \frac{1.86 \times 15.0 \times 1000}{0.34 \times 450} \)
\( M_2 = \frac{27900}{153} \)
\( \implies M_2 = 182.35 \text{ g mol}^{-1} \).
Hence, the molar mass of the substance is \( 182.35 \text{ g mol}^{-1} \).
In simple words: This question is solved by first defining mole fraction and ideal solution, and then using the freezing point depression of water to calculate the unknown molar mass.

Exam Tip: Be sure to write the formula for mole fraction clearly in the definition part to earn maximum presentation points.

Question. (a) Explain the following :
(i) Henry’s law about dissolution of a gas in a liquid
(ii) Boiling point elevation constant for a solvent
(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution? (Kb for water = 0.512 K kg mo^–1)
Answer:
(a) Explanations:
(i) Henry's law: The amount of gas that dissolves in a given volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with the liquid at a constant temperature.
(ii) Boiling point elevation constant (\( K_b \)): The increase in boiling point of a pure solvent when one mole of a non-volatile solute is dissolved in one kilogram of the solvent (i.e., a 1 molal solution). It is also known as the ebullioscopic constant.

(b) Numerical Calculation:
Given: solvent mass (water) \( w_1 = 500 \text{ g} \), constant \( K_b = 0.512 \text{ K kg mol}^{-1} \), boiling point of solution \( T_b = 100.42^\circ\text{C} \), and boiling point of pure water \( T_b^\circ = 100.00^\circ\text{C} \).
Calculate the boiling point elevation \( \Delta T_b \):
\( \Delta T_b = T_b - T_b^\circ = 100.42 - 100.00 = 0.42^\circ\text{C} \) (or \( 0.42 \text{ K} \)).
The molecular mass of glycerol (\( \text{C}_3\text{H}_8\text{O}_3 \)) is \( 3 \times 12 + 8 \times 1 + 3 \times 16 = 92 \text{ g mol}^{-1} \).
Using the boiling point elevation relation:
\( \Delta T_b = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1} \)
\( 0.42 = 0.512 \times \frac{w_2 \times 1000}{92 \times 500} \)
\( w_2 = \frac{0.42 \times 92 \times 500}{0.512 \times 1000} \)
\( w_2 = \frac{19320}{512} \)
\( \implies w_2 = 37.7 \text{ g} \).
Hence, the mass of glycerol dissolved in the solution is \( 37.7 \text{ g} \).
In simple words: This question requires us to define the concepts of gas solubility and boiling constants, then use boiling elevation data to find the exact mass of glycerol dissolved in water.

Exam Tip: Ensure that your definitions are accurate, and remember to show the calculation of glycerol's molar mass (92) step-by-step.

Question. (a) State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (\( K_f \) for benzene = 5.12 K kg mol\(^{-1}\))
Answer:
(a) Raoult’s law: For a solution of volatile liquids, the partial vapour pressure of each individual component in the mixture is directly proportional to its mole fraction present in that solution.
\[ P_A = P_A^\circ \times x_A \] Non-ideal solutions show deviations from Raoult’s law:
(i) Positive deviation from Raoult’s law: This occurs when the total vapour pressure of a mixture is greater than the predicted ideal vapour pressure at the same concentration. This behavior is called positive deviation. Example: Mixtures of ethanol and cyclohexane, or acetone and carbon disulphide.
(ii) Negative deviation from Raoult’s law: This happens when the overall vapour pressure is lower than the expected ideal value. This phenomenon is termed negative deviation. Example: Mixtures of chloroform and benzene, or chloroform and diethylether.

Special Case of Henry's Law:
According to Raoult’s law, the partial pressure is given by: \( P_A = P_A^\circ \times x_A \)
According to Henry’s law, the partial pressure is given by: \( P_A = K_H \times x_A \)
By comparing both expressions, it is evident that both laws are identical in form and only differ in their proportionality constants (\( P_A^\circ \) and \( K_H \)). Therefore, Raoult's law behaves as a special case of Henry's law where \( K_H \) becomes equal to \( P_A^\circ \).

(b) We utilize the formula:
\[ M_2 = \frac{1000 \times K_f \times w_2}{w_1 \times \Delta T_f} \]
Given values:
\( w_2 = 1.0\text{ g} \)
\( w_1 = 50\text{ g} \)
\( \Delta T_f = 0.40\text{ K} \)
\( K_f = 5.12\text{ K kg mol}^{-1} \)

Substituting these values into the formula:
\[ M_2 = \frac{1000 \times 5.12 \times 1.0}{50 \times 0.40} \]

\( \implies M_2 = 256\text{ g mol}^{-1} \)
Therefore, the molar mass of the non-electrolyte solute is \( 256\text{ g mol}^{-1} \).
In simple words: Raoult's law states that the pressure of a liquid component is related to its amount in the mixture. Positive deviation means the pressure is higher than expected, while negative means it is lower. Using the formula for freezing point depression, we calculated the molecular weight of the solute to be 256.

Exam Tip: When explaining deviations, always state the molecular interactions (like A-B vs A-A/B-B forces) to secure maximum marks. Make sure to clearly state units for molar mass in numerical parts.

Question. (a) Define the following terms :
(i) Ideal solution (ii) Azeotrope (iii) Osmotic pressure
(b) A solution of glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) in water is labelled as 10% by weight. What would be the molality of the solution?
(Molar mass of glucose = 180 g mol\(^{-1}\))
Answer:
(a) (i) Ideal solution: An ideal solution is one that strictly obeys Raoult's law over the entire range of concentrations, where the molecular forces between different components are identical in strength to those in the pure liquids.
(ii) Azeotrope: This is a special liquid mixture that boils at a constant temperature without any change in its composition during distillation, behaving like a single pure liquid.
(iii) Osmotic pressure: The minimum external pressure that must be applied to the solution to stop the flow of pure solvent across a semi-permeable membrane.

(b) A 10% by weight solution of glucose contains 10 g of glucose dissolved in 100 g of solution.
Mass of solute (glucose) = \( 10\text{ g} \)
Mass of solvent (water) = \( 100\text{ g} - 10\text{ g} = 90\text{ g} = 0.090\text{ kg} \)
Molar mass of glucose = \( 180\text{ g mol}^{-1} \)

Number of moles of glucose = \( \frac{10}{180} = 0.0556\text{ mol} \)

Molality (\( m \)) is calculated as:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]

\[ m = \frac{0.0556\text{ mol}}{0.090\text{ kg}} \]

\( \implies m \approx 0.617\text{ m} \)
The molality of the glucose solution is \( 0.617\text{ m} \).
In simple words: Ideal solutions follow Raoult's law, azeotropes boil at a single temperature without changing composition, and osmotic pressure stops solvent movement. A 10% glucose solution means we have 10 grams of sugar in 90 grams of water, which works out to a molality of 0.617.

Exam Tip: For percentage by weight problems, remember that the mass of the solvent is the total mass of the solution minus the mass of the solute. Do not mistake 100 g as the solvent mass.

Question. (a) The vapour pressures of benzene and toluene at 293 K are 75 mm Hg and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapour pressures are in equilibrium with the liquid mixture at this temperature.
(b) What is meant by +ve and –ve deviations from Raoult’s law and how is the sign of \( \Delta H_{\text{solution}} \) related to +ve and –ve deviations from Raoult’s law?
Answer:
(a) Given details:
Mass of benzene = \( 23.4\text{ g} \)
Molar mass of benzene (\( \text{C}_6\text{H}_6 \)) = \( 12 \times 6 + 6 = 78\text{ g mol}^{-1} \)
Mass of toluene = \( 64.4\text{ g} \)
Molar mass of toluene (\( \text{C}_6\text{H}_5\text{CH}_3 \)) = \( 12 \times 7 + 8 = 92\text{ g mol}^{-1} \)

Moles of benzene (\( n_B \)) = \( \frac{23.4}{78} = 0.3\text{ mol} \)
Moles of toluene (\( n_T \)) = \( \frac{64.4}{92} = 0.7\text{ mol} \)
Total moles = \( 0.3 + 0.7 = 1.0\text{ mol} \)

Mole fraction of benzene in liquid phase (\( x_B \)) = \( \frac{0.3}{1.0} = 0.3 \)
Mole fraction of toluene in liquid phase (\( x_T \)) = \( \frac{0.7}{1.0} = 0.7 \)

Partial vapour pressure of benzene (\( P_B \)) = \( x_B \times P_B^\circ = 0.3 \times 75 = 22.5\text{ mm Hg} \)
Partial vapour pressure of toluene (\( P_T \)) = \( x_T \times P_T^\circ = 0.7 \times 22 = 15.4\text{ mm Hg} \)
Total vapour pressure (\( P_{\text{total}} \)) = \( 22.5 + 15.4 = 37.9\text{ mm Hg} \)

Mole fraction of benzene in vapour phase (\( y_B \)) = \( \frac{P_B}{P_{\text{total}}} \)
\[ y_B = \frac{22.5}{37.9} = 0.59 \]

(b) Positive and Negative Deviations:
If the total vapor pressure of a mixture is higher than expected, it is a positive deviation. If it is lower, it shows a negative deviation.
For positive deviation, \( \Delta_{\text{mix}}H = +\text{ve} \) (endothermic process).
For negative deviation, \( \Delta_{\text{mix}}H = -\text{ve} \) (exothermic process).
In simple words: First find the moles of benzene and toluene to determine their pressures. By comparing the individual pressure of benzene to the total pressure, we get its mole fraction in the vapour as 0.59. Positive deviations absorb heat (positive change), while negative deviations release heat (negative change).

Exam Tip: Dalton's Law formula is used for calculating mole fractions in the vapour phase. Always clearly distinguish between liquid mole fractions (\( x \)) and vapour mole fractions (\( y \)) in your working.

Question. (a) A 5% solution (by mass) of cane-sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K.
[Molecular masses : Glucose \( \text{C}_6\text{H}_{12}\text{O}_6 \) : 180 amu; Cane-sugar \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) : 342 amu]
(b) State Henry’s law and mention two of its important applications.
Answer:
(a) For Cane-Sugar Solution:
A 5% solution by mass means \( 5\text{ g} \) of cane-sugar is dissolved in \( 95\text{ g} \) of water.
Molality of sugar solution (\( m_1 \)) = \( \frac{5 \times 1000}{342 \times 95} = 0.154\text{ mol kg}^{-1} \)
Depression in freezing point (\( \Delta T_f \)) = \( 273.15\text{ K} - 271\text{ K} = 2.15\text{ K} \)
Since \( \Delta T_f = K_f \times m \), we find:
\[ K_f = \frac{\Delta T_f}{m_1} = \frac{2.15}{0.154} = 13.96\text{ K kg mol}^{-1} \]

For Glucose Solution:
A 5% solution by mass of glucose means \( 5\text{ g} \) of glucose in \( 95\text{ g} \) of water.
Molality of glucose solution (\( m_2 \)) = \( \frac{5 \times 1000}{180 \times 95} = 0.292\text{ mol kg}^{-1} \)
Using the value of \( K_f \):
\[ \Delta T_f = K_f \times m_2 = 13.96 \times 0.292 \approx 4.08\text{ K} \]
Therefore, the freezing point of the glucose solution is:
\[ T_f = 273.15\text{ K} - 4.08\text{ K} = 269.07\text{ K} \]

(b) Henry’s Law: This law states that the mass or solubility of a gas dissolved in a liquid at a constant temperature is directly proportional to the partial pressure of that gas above the liquid surface.
Applications:
(i) Carbon dioxide gas is sealed under high pressure in soft drinks and soda bottles to increase its solubility.
(ii) Deep-sea divers use oxygen tanks diluted with helium to prevent the painful and dangerous condition known as "bends."
In simple words: We find the freezing constant using the sugar solution, then apply it to the glucose solution to calculate its freezing point as 269.07 K. Henry's law states that higher gas pressure increases how much gas dissolves in a liquid.

Exam Tip: Be precise with the mass of solvent. A 5% solution contains 95 g of water, not 100 g. Working with the correct mass of solvent prevents errors in final decimal values.

Question. (a) Define the following terms :
(i) Molarity (ii) Molal elevation constant (\( K_b \))
(b) A solution containing 15 g urea (molar mass = 60 g mol\(^{-1}\)) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water. Calculate the mass of glucose present in one litre of its solution.
Answer:
(a) (i) Molarity: This refers to the total number of moles of a solute that are dissolved in exactly one litre of solution.
(ii) Molal elevation constant (\( K_b \)): This is defined as the increase in boiling point observed when one mole of a non-volatile solute is dissolved in one kilogram of solvent.

(b) Since both solutions are isotonic, they must have the exact same molar concentration (osmotic pressure \( \Pi_1 = \Pi_2 \)).
Concentration of urea solution = \( \frac{\text{Mass of urea}}{\text{Molar mass of urea}} = \frac{15}{60}\text{ mol L}^{-1} \)
Concentration of glucose solution = \( \frac{\text{Mass of glucose } (w)}{\text{Molar mass of glucose}} = \frac{w}{180}\text{ mol L}^{-1} \)

Equating both concentrations:
\[ \frac{w}{180} = \frac{15}{60} \]

\( \implies w = \frac{15 \times 180}{60} \)

\( \implies w = 45\text{ g} \)
Thus, \( 45\text{ g} \) of glucose is present in one litre of the solution.
In simple words: Molarity measures moles per litre, and the molal elevation constant shows how much boiling point rises with a unit of concentration. Isotonic means equal concentrations. Since urea has a concentration of 15/60, glucose must match this, giving us 45 grams.

Exam Tip: For isotonic solutions, always start by equating the molarities of the two solutes (\( C_1 = C_2 \)). This simple step makes solving any isotonic problem straightforward.

Question. (a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution? (Density of solution = 1.2 g mL\(^{-1}\))
Answer:
(a) A mixture of ethanol and acetone displays positive deviation. In pure ethanol, strong hydrogen bonds exist between molecules. When acetone is introduced, its molecules fit into spaces between ethanol molecules and break some of these hydrogen bonds, which weakens the total intermolecular interactions.

(b) A 10% by mass glucose solution means \( 10\text{ g} \) of glucose is present in \( 100\text{ g} \) of solution.
Mass of glucose = \( 10\text{ g} \)
Mass of water = \( 100\text{ g} - 10\text{ g} = 90\text{ g} = 0.090\text{ kg} \)

Calculation of Molality:
\[ \text{Molality } (m) = \frac{\text{Mass of solute} \times 1000}{\text{Molar mass} \times \text{Mass of solvent in grams}} \]
\[ m = \frac{10 \times 1000}{180 \times 90} = 0.617\text{ m} \]

Calculation of Molarity:
Volume of \( 100\text{ g} \) of solution = \( \frac{\text{Mass}}{\text{Density}} = \frac{100\text{ g}}{1.2\text{ g mL}^{-1}} = 83.33\text{ mL} = 0.0833\text{ L} \)
Molarity (\( M \)) is given by:
\[ M = \frac{\text{Moles of glucose}}{\text{Volume of solution in L}} \]
\[ M = \frac{10 / 180}{0.0833} \]

\( \implies M \approx 0.67\text{ M} \)
Thus, the molality is \( 0.617\text{ m} \) and the molarity is \( 0.67\text{ M} \).
In simple words: Mixing ethanol and acetone breaks hydrogen bonds, leading to a positive deviation. Using the mass details, the concentration is 0.617 m for molality. Dividing moles by the solution volume (calculated from density) gives us a molarity of 0.67 M.

Exam Tip: Be careful with density. Molarity requires the volume of the entire solution, which you must calculate by dividing the solution mass (100 g) by its density.

Question. (a) What is van't Hoff factor? What types of values can it have if in forming the solution, the solute molecules undergo (i) Dissociation? (ii) Association?
(b) How many mL of a 0.1 M HCl solution are required to react completely with 1 g of a mixture of \( \text{Na}_2\text{CO}_3 \) and \( \text{NaHCO}_3 \) containing equimolar amounts of both? (Molar mass : \( \text{Na}_2\text{CO}_3 \) = 106 g, \( \text{NaHCO}_3 \) = 84 g)
Answer:
(a) The van't Hoff factor (\( i \)) is the ratio of the experimental (observed) value of a colligative property to its theoretically calculated value.
(i) Dissociation: \( i > 1 \) (since the total number of particles increases).
(ii) Association: \( i < 1 \) (since the total number of particles decreases).

(b) Let the mass of \( \text{Na}_2\text{CO}_3 \) in the mixture be \( x\text{ g} \).
Then, the mass of \( \text{NaHCO}_3 \) will be \( (1 - x)\text{ g} \).
Since the mixture contains equimolar amounts of both compounds:
\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \text{Moles of } \text{NaHCO}_3 \]
\[ \frac{x}{106} = \frac{1 - x}{84} \]
\[ 84x = 106(1 - x) \]
\[ 84x = 106 - 106x \]
\[ 190x = 106 \]
\[ x \approx 0.558\text{ g} \]
Mass of \( \text{Na}_2\text{CO}_3 = 0.558\text{ g} \), and mass of \( \text{NaHCO}_3 = 1 - 0.558 = 0.442\text{ g} \).

Moles of each component:
\[ \text{Moles} = \frac{0.558}{106} \approx 0.00526\text{ mol} \]

The chemical equations for the reactions with HCl are:
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]

Total moles of HCl required:
\[ \text{Moles of HCl} = 2 \times \text{Moles of } \text{Na}_2\text{CO}_3 + 1 \times \text{Moles of } \text{NaHCO}_3 \]
\[ \text{Moles of HCl} = 2(0.00526) + 0.00526 = 3 \times 0.00526 = 0.01578\text{ mol} \]

We now find the volume of \( 0.1\text{ M} \) HCl needed:
\[ \text{Volume of HCl in mL} = \frac{\text{Moles}}{\text{Molarity}} \times 1000 \]
\[ \text{Volume} = \frac{0.01578}{0.1} \times 1000 = 157.8\text{ mL} \]
Therefore, \( 157.8\text{ mL} \) of HCl is required.
In simple words: The van't Hoff factor is greater than 1 when molecules split and less than 1 when they join. For the mixture, we use algebra to find that we have 0.00526 moles of both compounds. In total, they need 0.01578 moles of acid, which equals 157.8 mL of the solution.

Exam Tip: Pay attention to the stoichiometry in the reactions. Remember that one mole of carbonate reacts with two moles of acid, while bicarbonate only reacts with one mole of acid.

Question. (a) Define (i) Mole fraction (ii) Molality (iii) Raoult's law
(b) Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber's salt, \( \text{Na}_2\text{SO}_4\cdot10\text{H}_2\text{O} \), in 0.100 kg of water. (\( K_f \) for water = 1.86 K kg mol\(^{-1}\), Atomic masses : Na = 23, S = 32, O = 16, H = 1)
Answer:
(a) (i) Mole fraction: This is the ratio of the number of moles of a single component to the total number of moles of all substances in a mixture.
(ii) Molality: The number of moles of solute dissolved per kilogram of solvent.
(iii) Raoult's law: For solutions of volatile components, the partial vapour pressure of each component is directly proportional to its mole fraction in the liquid phase.

(b) Molar mass of Glauber's salt (\( \text{Na}_2\text{SO}_4\cdot10\text{H}_2\text{O} \)):
\[ M = 2(23) + 32 + 4(16) + 10(18) = 46 + 32 + 64 + 180 = 322\text{ g mol}^{-1} \cancel{} \]
Moles of solute = \( \frac{6.00\text{ g}}{322\text{ g mol}^{-1}} = 0.0186\text{ mol} \)
Mass of solvent (water) = \( 0.100\text{ kg} \)
Molality (\( m \)) = \( \frac{0.0186\text{ mol}}{0.100\text{ kg}} = 0.186\text{ m} \)

Since Glauber's salt is an ionic compound, it completely dissociates in solution:
\[ \text{Na}_2\text{SO}_4\cdot10\text{H}_2\text{O} \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-} + 10\text{H}_2\text{O} \]
The total number of ions produced per formula unit is 3, so the van't Hoff factor \( i = 3 \).

Using the freezing point depression formula:
\[ \Delta T_f = i \times K_f \times m \]
\[ \Delta T_f = 3 \times 1.86 \times 0.186 \approx 1.04\text{ K} \]

Therefore, the freezing point of the solution (\( T_f \)) is:
\[ T_f = 273.15\text{ K} - 1.04\text{ K} = 272.11\text{ K} \text{ (or } -1.04^\circ\text{C)} \]
In simple words: Using the molecular mass of Glauber's salt (322 g/mol), we find the concentration is 0.186 m. Because the salt breaks up into 3 ions, we multiply the freezing point reduction by 3, lowering the freezing point of water to -1.04 degrees Celsius.

Exam Tip: Don't forget to include the water of crystallization (\( 10\text{H}_2\text{O} \)) when calculating the molar mass of Glauber's salt. Forgetting this is a very common mistake.

Question. (a) Calculate the freezing point of solution when 1.9 g of \( \text{MgCl}_2 \) (M = 95 g mol\(^{-1}\)) was dissolved in 50 g of water, assuming \( \text{MgCl}_2 \) undergoes complete ionization. (\( K_f \) for water = 1.86 K kg mol\(^{-1}\))
(b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution?
Answer:
(a) Given values:
Mass of \( \text{MgCl}_2 \) (\( w_2 \)) = \( 1.9\text{ g} \)
Molar mass of \( \text{MgCl}_2 \) (\( M_2 \)) = \( 95\text{ g mol}^{-1} \)
Mass of water (\( w_1 \)) = \( 50\text{ g} \)
\( K_f \) for water = \( 1.86\text{ K kg mol}^{-1} \)

Since \( \text{MgCl}_2 \) undergoes complete ionization:
\[ \text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^- \]
The total number of ions produced is 3, so \( i = 3 \).

Using the formula:
\[ \Delta T_f = i \times K_f \times m = i \times K_f \times \frac{w_2 \times 1000}{M_2 \times w_1} \]
\[ \Delta T_f = 3 \times 1.86 \times \frac{1.9 \times 1000}{95 \times 50} \]
\[ \Delta T_f = 5.58 \times 0.4 = 2.232\text{ K} \]

Freezing point of the solution (\( T_f \)):
\[ T_f = 273.15\text{ K} - 2.23\text{ K} = 270.92\text{ K} \text{ (or } -2.232^\circ\text{C)} \]

(b) (i) 2 M glucose has a higher boiling point. Elevation of boiling point is a colligative property that is determined by the total number of particles in solution. The concentration of solute particles in a 2 M solution is twice as high as in a 1 M solution.
(ii) When the applied external pressure exceeds the osmotic pressure, solvent molecules begin to flow from the concentrated solution back into the pure solvent through the semipermeable membrane. This process is called reverse osmosis.
In simple words: Since magnesium chloride splits into 3 parts, the temperature drops by 2.23 degrees, making the freezing point 270.92 K. A 2 M sugar solution has more particles, raising its boiling point higher. Pushing harder than the osmotic pressure causes reverse osmosis.

Exam Tip: Clearly show the ionization equation for ionic salts to explain why the van't Hoff factor (\( i \)) has a specific value. This shows the examiner you understand the chemistry behind the math.

Question. (a) When 2.56 g of sulphur was dissolved in 100 g of \( \text{CS}_2 \), the freezing point lowered by 0.383 K. Calculate the formula of sulphur (\( \text{S}_x \)). (\( K_f \) for \( \text{CS}_2 \) = 3.83 K kg mol\(^{-1}\), Atomic mass of Sulphur = 32 g mol\(^{-1}\))
(b) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?
Answer:
(a) Given values:
Mass of solute (sulphur) \( w_b = 2.56\text{ g} \)
Mass of solvent (\( \text{CS}_2 \)) \( w_a = 100\text{ g} \)
\( \Delta T_f = 0.383\text{ K} \)
\( K_f = 3.83\text{ K kg mol}^{-1} \)

Using the formula:
\[ M_b = \frac{K_f \times w_b \times 1000}{\Delta T_f \times w_a} \]
\[ M_b = \frac{3.83 \times 2.56 \times 1000}{0.383 \times 100} \]
\[ M_b = 256\text{ g mol}^{-1} \]

To find the number of atoms (\( x \)) in a molecule of sulphur:
\[ x = \frac{\text{Molar mass of sulphur molecule}}{\text{Atomic mass of sulphur}} \]
\[ x = \frac{256}{32} = 8 \]
Therefore, the molecular formula of sulphur is \( \text{S}_8 \).

(b) (i) If red blood cells are placed in a 1.2% NaCl solution (which is hypertonic), water will flow out of the cells into the outer solution, causing the cells to shrink.
(ii) If red blood cells are placed in a 0.4% NaCl solution (which is hypotonic), water will enter the cells, causing them to swell and potentially burst.
In simple words: The calculations show that the molecular weight of sulphur is 256, which means 8 atoms are linked together as S8. Since 1.2% salt is highly concentrated, water leaves blood cells and they shrink. In a weaker 0.4% solution, water enters the cells, making them swell.

Exam Tip: For cells in solution, always use the terms 'hypertonic' and 'hypotonic' to explain the direction of osmosis. This makes your answers look professional and scientifically accurate.

Question. (a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given : (Molar mass of sucrose = 342 g mol\(^{-1}\)) (Molar mass of glucose = 180 g mol\(^{-1}\))
(b) Define the following terms :
(i) Molality (m) (ii) Abnormal molar mass
Answer:
(a) For both solutions, a 10% solution by mass means \( 10\text{ g} \) of solute is dissolved in \( 90\text{ g} \) of water (solvent).

Depression in freezing point for sucrose solution:
\[ \Delta T_f(\text{sucrose}) = 273.15\text{ K} - 269.15\text{ K} = 4.00\text{ K} \]

Using the formula \( \Delta T_f = K_f \times m \):
\[ \Delta T_f(\text{sucrose}) = K_f \times \frac{w_{\text{sucrose}} \times 1000}{M_{\text{sucrose}} \times w_{\text{water}}} \]
\[ 4.00 = K_f \times \frac{10 \times 1000}{342 \times 90} \quad \text{--- (Equation 1)} \]

For glucose solution:
\[ \Delta T_f(\text{glucose}) = K_f \times \frac{10 \times 1000}{180 \times 90} \quad \text{--- (Equation 2)} \]

Dividing Equation 2 by Equation 1:
\[ \frac{\Delta T_f(\text{glucose})}{4.00} = \frac{342}{180} \]
\[ \Delta T_f(\text{glucose}) = 4.00 \times \frac{342}{180} = 7.60\text{ K} \]

Therefore, the freezing point of the glucose solution is:
\[ T_f = 273.15\text{ K} - 7.60\text{ K} = 265.55\text{ K} \]

(b) (i) Molality (m): This indicates the number of moles of solute dissolved in exactly one kilogram of solvent.
(ii) Abnormal molar mass: When the calculated molar mass of a substance using colligative properties is higher or lower than its actual chemical formula mass, it is called an abnormal molar mass. This happens due to the association or dissociation of solute particles in the solution.
In simple words: Since both solutions have the same concentration by weight, the ratio of their freezing point changes depends only on their molecular weights. This gives a drop of 7.60 K for glucose, resulting in a freezing point of 265.55 K.

Exam Tip: Instead of calculating \( K_f \) separately, dividing the two equations is much quicker and avoids rounding errors during long calculations.

Question. (a) 30 g of urea (M = 60 g mol\(^{-1}\)) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.
Answer:
(a) Given details:
Mass of urea (\( w_2 \)) = \( 30\text{ g} \), Molar mass (\( M_2 \)) = \( 60\text{ g mol}^{-1} \)
Mass of water (\( w_1 \)) = \( 846\text{ g} \), Molar mass (\( M_1 \)) = \( 18\text{ g mol}^{-1} \)
Vapour pressure of pure water (\( P^\circ \)) = \( 23.8\text{ mm Hg} \)

Moles of urea (\( n_2 \)) = \( \frac{30}{60} = 0.5\text{ mol} \)
Moles of water (\( n_1 \)) = \( \frac{846}{18} = 47\text{ mol} \)

Using Raoult's Law for relative lowering of vapour pressure:
\[ \frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \]
\[ \frac{23.8 - P_s}{23.8} = \frac{0.5}{47 + 0.5} = \frac{0.5}{47.5} \]
\[ 23.8 - P_s = 23.8 \times 0.01053 \]
\[ 23.8 - P_s \approx 0.25\text{ mm Hg} \]

\( \implies P_s = 23.8 - 0.25 = 23.55\text{ mm Hg} \)
The vapour pressure of water in this solution is \( 23.55\text{ mm Hg} \).

(b) Differences between Ideal and Non-Ideal Solutions:

In simple words: By using the mole ratio of urea to water, we calculate how much the vapour pressure drops. It decreases by 0.25 mm Hg, leaving the solution's vapor pressure at 23.55 mm Hg. Ideal mixtures show no heat or volume changes, unlike non-ideal ones.

Exam Tip: For differences, always use a table to compare. It makes the distinction extremely clear and helps you score full marks.

Question. (a) Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases.
(b) Henry's law constant for \( \text{CO}_2 \) in water is \( 1.67 \times 10^8\text{ Pa} \) at 298 K. Calculate the number of moles of \( \text{CO}_2 \) in 500 ml of soda water when packed under \( 2.53 \times 10^5\text{ Pa} \) at the same temperature.
Answer:
(a) Glucose is a non-volatile solute. Adding it to water lowers the vapour pressure of water. Since boiling occurs when the vapour pressure equals atmospheric pressure, a higher temperature is required to reach this state, causing the boiling point to rise.

(b) Given details:
\( K_H = 1.67 \times 10^8\text{ Pa} \)
\( P_{\text{CO}_2} = 2.53 \times 10^5\text{ Pa} \)
Using Henry's law:
\[ P_{\text{CO}_2} = K_H \times x_{\text{CO}_2} \]
\[ x_{\text{CO}_2} = \frac{P_{\text{CO}_2}}{K_H} = \frac{2.53 \times 10^5}{1.67 \times 10^8} = 1.515 \times 10^{-3} \]

Now, let's find the moles of water present in \( 500\text{ mL} \) of soda water (assuming density of soda water is \( 1\text{ g mL}^{-1} \)):
Mass of water = \( 500\text{ g} \)
\[ \text{Moles of water } (n_{\text{H}_2\text{O}}) = \frac{500}{18} \approx 27.78\text{ mol} \cancel{} \]
Since \( n_{\text{CO}_2} \ll n_{\text{H}_2\text{O}} \), we can write the mole fraction as:
\[ x_{\text{CO}_2} \approx \frac{n_{\text{CO}_2}}{n_{\text{H}_2\text{O}}} \]
\[ 1.515 \times 10^{-3} = \frac{n_{\text{CO}_2}}{27.78} \]

\( \implies n_{\text{CO}_2} = 1.515 \times 10^{-3} \times 27.78 \approx 0.042\text{ mol} \)
Therefore, the number of moles of \( \text{CO}_2 \) in the soda water is \( 0.042\text{ mol} \).
In simple words: Sugar doesn't evaporate, so putting it in water lowers the pressure of the water vapour, requiring more heat to boil. Using Henry's law, we calculated that the concentration of carbon dioxide in the soda water is 0.042 moles.

Exam Tip: In mole fraction calculations where the solute concentration is very low, make the approximation \( n_{\text{solute}} + n_{\text{solvent}} \approx n_{\text{solvent}} \). This simplifies your calculations significantly.

Question. (a) Define the following terms :
(i) Ideal solution (ii) Osmotic pressure
(b) Calculate the boiling point elevation for a solution prepared by adding 10 g \( \text{CaCl}_2 \) to 200 g of water, assuming that \( \text{CaCl}_2 \) is completely dissociated. (\( K_b \) for water = 0.512 K kg mol\(^{-1}\); Molar mass of \( \text{CaCl}_2 \) = 111 g mol\(^{-1}\))
Answer:
(a) (i) Ideal solution: A solution that obeys Raoult's law across the entire concentration range and shows no change in enthalpy or volume upon mixing.
(ii) Osmotic pressure: The precise excess pressure that must be applied to a solution to prevent the inward flow of solvent through a semipermeable membrane.

(b) Given details:
Mass of solute (\( \text{CaCl}_2 \)) \( w_2 = 10\text{ g} \)
Mass of solvent (water) \( w_1 = 200\text{ g} \)
\( K_b = 0.512\text{ K kg mol}^{-1} \)
Molar mass of \( \text{CaCl}_2 \) \( M_2 = 111\text{ g mol}^{-1} \)

Since \( \text{CaCl}_2 \) is completely dissociated:
\[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \]
The total number of ions produced is 3, so \( i = 3 \).

Using the boiling point elevation formula with the van't Hoff factor:
\[ \Delta T_b = i \times K_b \times m = i \times K_b \times \frac{w_2 \times 1000}{M_2 \times w_1} \]
\[ \Delta T_b = 3 \times 0.512 \times \frac{10 \times 1000}{111 \times 200} \]
\[ \Delta T_b = 1.536 \times 0.45 \approx 0.69\text{ K} \]
Therefore, the boiling point elevation is \( 0.69\text{ K} \).
In simple words: Ideal solutions follow the rules perfectly, and osmotic pressure blocks solvent movement. Since calcium chloride releases 3 ions when dissolved, the elevation value is multiplied by 3, which increases the boiling temperature by 0.69 K.

Exam Tip: Always check if a solute dissociates or associates. For strong electrolytes like \( \text{CaCl}_2 \), you must include the van't Hoff factor (\( i \)) in the formula, otherwise your final calculation will be incorrect.