RBSE Solutions Class 12 Chemistry Chapter 2 Solution

RBSE Class 12 Chemistry Chapter 2 Text Book Questions

RBSE Class 12 Chemistry Chapter 2 Multiple Choice Questions

 

Question 1. 4 g of NaOH is dissolved in 500 g of water. The concentration of solution will be
(a) 8 g/L
(b) 0.2.N.
(c) 0.2 m
(d) 0.2 M
Answer: (c) 0.2 m
In simple words: When 4 grams of NaOH is mixed into 500 grams of water, the solution's concentration is 0.2 molal (m). Molality measures the number of moles of solute per kilogram of solvent.

🎯 Exam Tip: Remember to distinguish between molarity (moles/liter of solution) and molality (moles/kilogram of solvent) as their definitions affect calculations.

 

Question 2. Which solution represents positive deviation from Raoult's law ?
(Options for this question are missing from the source content)
Answer: (c)
In simple words: A positive deviation from Raoult's law means the vapor pressure of the solution is higher than expected. This usually happens when the forces between different types of molecules in the solution are weaker than the forces between molecules of the same type.

🎯 Exam Tip: Positive deviation indicates weaker A-B interactions compared to A-A and B-B interactions, leading to easier escape of molecules into the vapor phase.

 

Question 3. The molarity of pure water is-
(a) 55.5 M
(b) 100 M
(c) 18 M
(d) IM
Answer: (a) 55.5 M
In simple words: The molarity of pure water is about 55.5 M. This means that in one liter of pure water, there are roughly 55.5 moles of water molecules. Water's density is approximately 1 g/mL.

🎯 Exam Tip: To calculate the molarity of water, assume 1 liter of water (1000g), then divide by its molecular mass (18 g/mol).

 

Question 4. Arrange the following 0.1 M solutions in increasing order of their boiling points
(a) NaCl
(b) MgCl\(_{2}\)
(c) Urea
(d) AlCl\(_{3}\)
(a) (i) < (ii) < (iii) < (iv).
(b) (ii) < (i) < (iii) < (iv)
(c) (iii) < (i) < (ii) < (iv)
(d) (iv) < (iii) < (ii) < (i)
Answer: (c) (iii) < (i) < (ii) < (iv)
In simple words: The boiling point of a solution depends on how many particles are dissolved in it. Urea (iii) makes 1 particle. NaCl (i) makes 2 particles (\( Na^+ \), \( Cl^- \)). MgCl\(_{2}\) (ii) makes 3 particles (\( Mg^{2+} \), 2\( Cl^- \)). AlCl\(_{3}\) (iv) makes 4 particles (\( Al^{3+} \), 3\( Cl^- \)). More particles mean a higher boiling point.

🎯 Exam Tip: Remember to count the total number of ions or particles each substance produces when dissolved in water to correctly predict the effect on colligative properties like boiling point elevation.

 

Question 5. This is a property of an ideal solution
(a) It obeys Raoult's law
(b) \( \Delta H_{mixing} \)
(c) \( \Delta V_{mixing} \)
(d) All of the options
Answer: (d) All of the options
In simple words: An ideal solution is one where the particles mix without any change in heat or volume. It also perfectly follows Raoult's law, meaning its vapor pressure behaves predictably. Ideal solutions have no interaction differences between components.

🎯 Exam Tip: For an ideal solution, the enthalpy of mixing (\( \Delta H_{mixing} \)) and the volume of mixing (\( \Delta V_{mixing} \)) are both zero, and it perfectly follows Raoult's law.

 

Question 6. On increasing temperature, vapour pressure of a substance
(a) always increases.
(b) decreases.
(c) does not depend on temperature.
(d) partially depends on temperature.
Answer: (a) always increases.
In simple words: When a substance gets hotter, its particles move faster and have more energy. This makes it easier for them to escape into the air above the liquid, so the vapor pressure always goes up. This is a fundamental concept in thermodynamics.

🎯 Exam Tip: Temperature provides kinetic energy to molecules, enabling more of them to overcome intermolecular forces and enter the vapor phase, thus increasing vapor pressure.

 

Question 7. The freezing point of 5% sugar solution will be
(Options for this question are missing from the source content)
Answer: (a)
In simple words: When sugar is added to water, it lowers the freezing point of the water. This means the solution will freeze at a temperature below 0°C because the sugar molecules interfere with the formation of ice crystals. The amount of depression depends on the concentration.

🎯 Exam Tip: Freezing point depression is a colligative property, meaning it depends on the number of solute particles, not their identity. Use \( \Delta T_f = K_f m \) for calculations.

 

Question 8. On increasing temperature, the solubility of H\(_{2}\) gas in water
(a) increases
(b) decreases
(c) remains unchanged
(d) None of the options
Answer: (a) increases
In simple words: As water gets hotter, hydrogen gas becomes more soluble in it. This is because the gas molecules have more energy to mix with the water molecules. Most gases, however, become less soluble in liquids as temperature increases.

🎯 Exam Tip: Generally, the solubility of gases in liquids decreases with increasing temperature due to the exothermic nature of dissolution. However, hydrogen is an exception because its dissolution is endothermic at lower temperatures.

RBSE Class 12 Chemistry Chapter 2 Very Short Answer Type Questions

 

Question 1. Calculate molality of 10% (w/W) aqueous H\(_{2}\)SO\(_{4}\)
Answer:
Given: Weight of solute (H\(_{2}\)SO\(_{4}\)), \( W_B = 10 \text{ g} \)
Weight of solvent (water), \( W_A = 100 - 10 = 90 \text{ g} \)
Molecular weight of solute (H\(_{2}\)SO\(_{4}\)), \( M_B = 98 \text{ g mol}^{-1} \)

The formula for molality (m) is:
\[ m = \frac{W_B \times 1000}{M_B \times W_A} \]
Now we put in the values:
\[ m = \frac{10 \times 1000}{98 \times 90} \]
\[ m = \frac{10000}{8820} \]
\[ m = 1.134 \text{ mol kg}^{-1} \]
In simple words: To find the molality, we first figure out the mass of sulfuric acid (solute) and water (solvent). Then, we use the formula that divides the moles of solute by the mass of solvent in kilograms. This calculation shows the concentration is 1.134 molal.

🎯 Exam Tip: Always pay attention to whether a percentage is by mass (w/W), mass/volume (w/V), or volume/volume (v/v) to correctly determine the masses of solute and solvent.

 

Question 2. What is molarity? Write the effect of temperature on it.
Answer:
Molarity is the number of moles of solute dissolved per litre of the solution.
The formula for molarity (M) is:
\[ M = \frac{W_B \times 1000}{M_B \times V_{(sol.)}} \]
When temperature increases, the molarity of a solution decreases. This happens because increasing the temperature causes the volume of the solution to increase. Since molarity is moles per unit volume, a larger volume with the same number of moles results in lower molarity.
In simple words: Molarity tells us how many moles of a substance are in one liter of solution. If the temperature goes up, the solution gets bigger in volume, so the molarity goes down.

🎯 Exam Tip: Since molarity is volume-dependent, it changes with temperature. Molality, being mass-dependent, is a better choice for temperature-independent concentration measurements.

 

Question 3. What are the differences between diffusion and osmosis? Give an example of each. Represent the diffusion and osmosis by labelled diagrams.
Answer:
Differences between diffusion and osmosis:

Examples:
Diffusion: When potassium permanganate (purple crystals) is added to a container of water, the violet color spreads throughout the container over time. This is because the dye particles move from an area of high concentration to low concentration until evenly distributed.
Osmosis: If grapes are placed in a concentrated salt solution, they will shrink because water moves out of the grapes into the more concentrated solution. Conversely, if dried raisins or shrunken grapes are put into pure water, they swell as water moves into them.
In simple words: Diffusion is when things spread out on their own, like perfume in a room. Osmosis is similar but only water moves through a special filter (semi-permeable membrane) from a watery area to a less watery area.

🎯 Exam Tip: The key difference lies in the semi-permeable membrane and what moves: in osmosis, only solvent (usually water) moves; in diffusion, both solute and solvent particles can move.

 

Question 4. Does it advise to use ethylene glycol in radiators of cars in summer?
Answer:
No, it is not advised to use ethylene glycol in radiators of cars in summer. Ethylene glycol is primarily used as an antifreeze in winter. It works by lowering the freezing point of water, which prevents the radiator fluid from freezing in cold temperatures. In summer, the main concern is overheating, and while ethylene glycol does raise the boiling point of water, adding it when not strictly necessary can reduce the cooling efficiency compared to pure water or specialized coolants, and it is usually already present in antifreeze mixtures suitable for all seasons. Therefore, its primary benefit is for cold protection rather than hot weather performance alone.
In simple words: No, it's not advised to use ethylene glycol in car radiators only in summer. Its main job is to stop water from freezing in winter. For summer, other coolants or proper mixtures are better to prevent the engine from getting too hot.

🎯 Exam Tip: Ethylene glycol is used to prevent freezing (antifreeze) and also slightly raises the boiling point (antiboil), making it suitable for year-round use in mixed solutions, but its primary function is cold protection.

 

Question 5. Define reverse osmosis?
Answer:
Reverse osmosis is a process where a solvent is forced from a region of high solute concentration through a semi-permeable membrane to a region of low solute concentration. This happens when an external pressure, which is higher than the osmotic pressure, is applied to the concentrated solution side. Essentially, it reverses the natural direction of osmosis, which normally involves solvent moving from low to high solute concentration. This process is commonly used to purify water by removing salts and other impurities.
In simple words: Reverse osmosis is a way to clean water by pushing it through a special filter. This forces the water to leave behind salt and other bad stuff, making it pure.

🎯 Exam Tip: Reverse osmosis requires external pressure greater than the osmotic pressure to overcome the natural flow and push solvent from a high concentration to a low concentration side.

RBSE Class 12 Chemistry Chapter 2 Short Answer Type Questions

Effect of temperature on solubility of solid in liquid- There is following equilibrium between soluble solid and solution in a saturated solution. Insoluble solid + Solute – Solvent Solution \( \rightleftharpoons \) Solution ; \( \Delta H_{sol} = \pm x \text{ k cal} \)

According to Le-chatelier's principle if \( \Delta H > 0 \) (zero) i.e. on dissolving solute in solvent, energy is absorbed, hence on increasing temperature the solubility of solid solute increases. Example: \( NH_4 Cl, KCl, AgNO_3, NaNO_3, KI \) etc.

If \( \Delta H < 0 \) (zero) i.e. on dissolving solute in solvent energy is released, hence on increasing temperature the solubility of solid solute decreases. Example: \( NaOH, Li_2SO_4, (CH_3 COO)_3 Ca \) etc.

van't Hoff's factor

\((i) = \frac{\text{Normal (Calculated) Molecular Mass}}{\text{Abnormal (Observed) Molecular Mass}}\)

\(i = \frac{M_c}{M_o}\)

If observed colligative property < calculated value or \( M_o > M_c \) (or \( i < 1 \)), there is association. If observed colligative property > calculated value or \( M_o < M_c \) (or \( i > 1 \)), there is dissociation.

 

Question 2. The theoretical molecular mass and observed molecular mass of an ionic compound AB are 58.2 and 30 respectively. Calculate the van't Hoff factor and degree of dissociation.
Answer:
Given:
Calculated molecular mass of AB = 58.2
Observed molecular mass of AB = 30

Van't Hoff's factor (i) is calculated as:
\[ i = \frac{\text{Calculated molecular mass}}{\text{Observed molecular mass}} \]
\[ i = \frac{58.2}{30} \]
\[ i = 1.94 \]
The degree of dissociation \( (\alpha) \) for a compound like AB, which dissociates into two ions (A\(^+\) and B\(^-\)), means \( n = 2 \).
The formula for the degree of dissociation is:
\[ \alpha = \frac{i-1}{n-1} \]
Substitute the values:
\[ \alpha = \frac{1.94 - 1}{2 - 1} \]
\[ \alpha = \frac{0.94}{1} \]
\[ \alpha = 0.94 \]
In simple words: The van't Hoff factor helps us see how many particles a substance breaks into when dissolved. By comparing the expected mass to the actual measured mass, we found it breaks into almost two pieces (1.94). The degree of dissociation then tells us that 94% of the compound separates into ions in the solution.

🎯 Exam Tip: Remember that for dissociation, \( n \) represents the number of ions formed from one formula unit of the solute, which is crucial for calculating the degree of dissociation.

 

Question 3. What are the differences between diffusion and osmosis? Give an example of each. Represent the diffusion and osmosis by labelled digrams.
Answer:
Differences between diffusion and osmosis:

Examples:
Diffusion: When potassium permanganate (purple crystals) is added to a container of water, the violet color spreads throughout the container over time. This is because the dye particles move from an area of high concentration to low concentration until evenly distributed.
Osmosis: If grapes are placed in a concentrated salt solution, they will shrink because water moves out of the grapes into the more concentrated solution. Conversely, if dried raisins or shrunken grapes are put into pure water, they swell as water moves into them.
(Diagrams are not provided in the source content.)
In simple words: Diffusion is when things spread out on their own, like perfume in a room. Osmosis is similar but only water moves through a special filter (semi-permeable membrane) from a watery area to a less watery area.

🎯 Exam Tip: The key difference lies in the semi-permeable membrane and what moves: in osmosis, only solvent (usually water) moves; in diffusion, both solute and solvent particles can move.

 

Question 4. 0.2L aqueous solution of protein contains 1.26 g protein. The osmotic pressure of this solution is 2. 57 x 10\(^{-3}\) bar at 300 K. Calculate the molecular mass of protein.
Answer:
Given:
Volume 'V' = 0.2 L
Mass of solute 'W\(_{B}\)' = 1.26 g
Temperature 'T' = 300 K
Osmotic pressure '\(\pi\)' = \(2.57 \times 10^{-3}\) bar
Gas constant 'R' = 0.0821 L bar mol\(^{-1}\) K\(^{-1}\)

The formula for osmotic pressure is:
\[ \pi = \frac{W_B \times R \times T}{M_B \times V} \]
To find the molecular mass of protein (\( M_B \)), we rearrange the formula:
\[ M_B = \frac{W_B \times R \times T}{\pi \times V} \]
Now we substitute the given values:
\[ M_B = \frac{1.26 \text{ g} \times 0.0821 \text{ L bar mol}^{-1}\text{K}^{-1} \times 300 \text{ K}}{0.2 \text{ L} \times 2.57 \times 10^{-3} \text{ bar}} \]
\[ M_B = \frac{31.0266}{5.14 \times 10^{-4}} \]
\[ M_B = 60363.03 \text{ g/mol} \]
The molecular mass of the protein is approximately 60363.03 g/mol.
In simple words: We used the osmotic pressure formula, which links pressure, mass, gas constant, temperature, volume, and molecular mass. By plugging in the known values for the protein solution, we can find out how heavy one mole of the protein is.

🎯 Exam Tip: The osmotic pressure formula (\( \pi V = nRT \)) is often rearranged to solve for molecular mass, making it a valuable tool for determining the molecular mass of large molecules like proteins.

Elevation in boiling point is directly proportional to lowering in osmotic pressure.

\( \Delta T_b \propto \Delta P \)

According to Raoult's law, lowering in vapour pressure is directly proportional to mole fraction of solute, hence,
\( \Delta P \propto X_b \) (\( X_b \) = mole fraction of solute)

Hence,
\( \Delta T_b \propto X_B \)

or
\[ \Delta T_b = kX_B = k \frac{n_B}{n_A + n_B} \]
\[ = k \frac{W_B/M_B}{W_A/M_A + W_B/M_B} \]

For dilute solution \( \frac{W_B}{M_B} \ll \frac{W_A}{M_A} \) and hence \( \frac{W_B}{M_B} \) may be neglected in denominator as compared to \( \frac{W_A}{M_A} \).

\( \therefore \Delta T_b = k \frac{W_B/M_B}{W_A/M_A} = k \frac{n_B M_A}{W_A} \)

If \( W_A \) is the mass of solvent in kg, then \( \frac{n_B}{W_A} \) is equal to molality (m) of the solution.

\( \Delta T_b = K_f m \)

Here k and \( M_A \) are constants and hence their product i.e., \( K M_A \) is replaced by another constant \( K_b \).

\( \Delta T_b = K_b m \)

Where \( K_b \) is called boiling point-elevation constant or molal elevation constant or molal ebullioscopic constant.

 

Question 6. How can the molecular mass of non-volatile substance determined from the component of vapour pressure? Explain it.
Answer:
For non-volatile substances, the relative lowering in vapor pressure of the solution is equal to the mole fraction of the solute. This principle is described by Raoult's law.

Where,
\( n_B \) = Number of moles of solute
\( n_A \) = Number of moles of solvent
\( W_B \) = Mass of solute
\( W_A \) = Mass of solvent
\( M_B \) = Molecular mass of solute
\( M_A \) = Molecular mass of solvent

According to Raoult's law, the mole fraction of solute \( X_B \) is given by:
\[ X_B = \frac{P_A^\circ - P_A}{P_A^\circ} \]
Also, the mole fraction \( X_B \) can be expressed in terms of moles:
\[ X_B = \frac{n_B}{n_A + n_B} = \frac{W_B/M_B}{W_A/M_A + W_B/M_B} \]
For dilute solutions, the number of moles of solute (\( n_B \)) is much smaller than the number of moles of solvent (\( n_A \)). Therefore, \( n_A + n_B \) can be approximated as \( n_A \).
So, for dilute solutions:
\[ X_B \approx \frac{n_B}{n_A} = \frac{W_B/M_B}{W_A/M_A} = \frac{W_B \times M_A}{M_B \times W_A} \]
Equating the two expressions for \( X_B \):
\[ \frac{P_A^\circ - P_A}{P_A^\circ} = \frac{W_B \times M_A}{M_B \times W_A} \]
From this equation, we can rearrange to solve for the molecular mass of the non-volatile solute (\( M_B \)):
\[ M_B = \frac{W_B \times M_A \times P_A^\circ}{W_A \times (P_A^\circ - P_A)} \]
With the help of the above equation, we can calculate the value of the molecular mass of a non-volatile solute if all other factors in the equation are known. This method provides an accurate way to determine unknown molecular masses.
In simple words: We can find the molecular mass of a non-volatile substance by measuring how much it lowers the vapor pressure of a solvent. Using Raoult's law, we connect this lowering to the amount of substance present and then calculate its molecular mass.

🎯 Exam Tip: This method is particularly useful for non-volatile solutes because they do not contribute to the vapor pressure, making the observed lowering directly attributable to the solute's presence.

 

Question 7. Explain the factors affecting solubility of gases in liquid.
Answer:
The solubility of gases in liquids is affected by several factors:
1. Nature of gas: Gases that react with the solvent or ionize in solution are generally more soluble. For example, ammonia (NH\(_{3}\)), hydrogen chloride gas (HCl(g)), and sulfur dioxide (SO\(_{2}\)) are highly soluble in water because they react with it to form compounds like \( NH_4OH, HCl(l), \) and \( H_2SO_4 \).
2. Nature of solvent: The principle "like dissolves like" applies here. Polar gases (like HCl) are more soluble in polar solvents (like water), while non-polar gases are more soluble in non-polar solvents (like benzene).
3. Effect of Temperature: For most gases, increasing the temperature decreases their solubility in liquids. This is because the dissolution of gases in liquids is an exothermic process. According to Le Chatelier's principle, an increase in temperature shifts the equilibrium in the direction that absorbs heat, which is the reverse direction (gas coming out of solution). However, for a few gases like hydrogen at lower temperatures, the dissolution can be endothermic, and their solubility increases with temperature.
4. Effect of pressure: The solubility of gases is significantly influenced by pressure, as described by Henry's Law. Henry's Law states that "at constant temperature, the amount of gas soluble in a unit volume of solvent is directly proportional to the pressure created by the gas at equilibrium on the surface of the solvent." So, if the mass of gas is 'm' dissolved in a unit volume of solvent at pressure 'P', then \( m \propto P \), or \( m = kP \), where 'k' is Henry's constant.
In simple words: How much gas dissolves in a liquid depends on what kind of gas it is, what kind of liquid it is, the temperature, and the pressure. For example, cold, fizzy drinks have more gas because lower temperature and higher pressure help the gas stay dissolved.

🎯 Exam Tip: Remember Henry's Law for the effect of pressure and that for most gases, increasing temperature *decreases* solubility in liquids.

 

Question 8. Calculate the temperature at which the solution of 54g glucose present in 250g water will be freezed. (K\(_{f}\) =1.86 K kg mol\(^{-1}\))
Answer:
Given:
Mass of glucose (\( W_B \)) = 54 g
Mass of water (\( W_A \)) = 250 g
Molal depression constant (\( K_f \)) = 1.86 K kg mol\(^{-1}\)
Molecular mass of glucose (\( M_B \)) = 180 g/mol (C\(_{6}\)H\(_{12}\)O\(_{6}\))
Freezing point of pure water (\( T_f^\circ \)) = 273.15 K

First, calculate the depression in freezing point (\( \Delta T_f \)) using the formula:
\[ \Delta T_f = \frac{K_f \times W_B \times 1000}{M_B \times W_A} \]
Substitute the given values:
\[ \Delta T_f = \frac{1.86 \text{ K kg mol}^{-1} \times 54 \text{ g} \times 1000}{180 \text{ g/mol} \times 250 \text{ g}} \]
\[ \Delta T_f = \frac{100440}{45000} \]
\[ \Delta T_f = 2.232 \text{ K} \]
Now, calculate the freezing point of the solution (\( T_f \)):
\[ T_f = T_f^\circ - \Delta T_f \]
\[ T_f = 273.15 \text{ K} - 2.232 \text{ K} \]
\[ T_f = 270.918 \text{ K} \]
Rounding to two decimal places, the freezing point of the solution is approximately 270.92 K.
In simple words: We calculated how much the freezing point of water drops when glucose is added. We used a special formula with the given values for glucose and water. The result shows that the solution will freeze at a temperature lower than pure water, specifically at about 270.92 Kelvin.

🎯 Exam Tip: Ensure that the units are consistent (e.g., mass in grams, solvent mass in kilograms for \( K_f \) unit, temperature in Kelvin) when applying the freezing point depression formula.

 

Question 9. Draw a graph for an ideal solution made up by solute and solvent.
Answer:
The graph for an ideal solution, according to Raoult's law, shows the partial vapor pressures of each component and the total vapor pressure of the solution as a function of the mole fraction of the components. For an ideal binary solution of components A and B:
\[ P_A = X_A P_A^\circ \]
\[ P_B = X_B P_B^\circ \]
\[ P_{total} = P_A + P_B = X_A P_A^\circ + X_B P_B^\circ \]
The graph displays linear relationships for both partial pressures and the total pressure. The lines extend from the pure component vapor pressures when their mole fraction is 1, down to zero when their mole fraction is 0.

In simple words: This graph shows how the total air pressure above a mixed liquid changes as we change the amounts of the two liquids in it. Each liquid adds its own pressure, and the total is just the sum of these, following a simple straight-line rule.

🎯 Exam Tip: Ensure that the x-axis represents the mole fraction of one component (e.g., \( X_A \) varying from 0 to 1) and that the pure vapor pressures \( P_A^\circ \) and \( P_B^\circ \) are correctly marked on the y-axis.