CBSE Class 12 Chemistry Solutions Question Bank Set A

Question. Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in moles/lit will be:
(a) 0.33
(b) 0.666
(c) 0.0033
(d) 3
Answer. C

Question. The value of Henry’s Law constant is:
(a) larger for gases with higher solubility
(b) larger for gases with lower solubility
(c) constant for all gases
(d) not related to the solubility of gases
Answer. B

Question. The elevation in boiling point of 0.01 M BaCl₂ solution is about than that of 0.01 M solution of glucose.
(a) Same
(b) two times
(c) three times
(d) four times
Answer. C

Question. Considering the formation, breaking and strength of Hydrogen bond, predict which of the following mixtures will show a positive deviation from Roult’s law?
(a) Methanol and Acetone
(b) Chloroform and Acetone
(c) Nitric Acid and Water
(d) Phenol and Aniline
Answer. A

Question. If a molecule AB undergoes dimerization in Benzene, its Van’t Hoff factor is found to be 0.60. The degree of dissociation of AB is
(a) 20%
(b) 60%
(c) 80%
(d) 50%
Answer. C

Question. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is
(a) 3.28 mol kg^{– 1}
(b) 2.28 mol kg^{– 1}
(c) 0.44 mol kg^{– 1}
(d) 1.14 mol kg^{– 1}
Answer. B

Question. At certain temperature, a 5.12% solution of cane sugar is isotonic with a 0.9% solution of an unknown solute. The molar mass of solute is
(a) 60
(b) 46.67
(c) 120
(d) 90
Answer. A

Question. Which is not a colligative property?
(a) Osmotic pressure
(b) Lowering of vapour pressure
(c) Depression in freezing point
(d) Molal elevation constant
Answer. D

Question. 12g of Urea is dissolved in 1L of water and 68.4g sucrose is dissolved in 1L of water. Relative lowering of vapour pressure of Urea solution is:
(a) Greater than sucrose solution
(b) Less than sucrose solution
(c) Double that of sucrose solution
(d) Equal to that of sucrose solution
Answer. D

Question. Which of the following is dependent on temperature?
(a) Molality
(b) Molarity
(c) Mole Fraction
(d) Mass percentage
Answer. B

SHORT ANSWER QUESTIONS

Question. How does sprinkling of salt help in clearing the snow-covered roads in hilly areas? Explain the phenomenon involved in the process.
Answer. The phenomenon involved in the melting of snow in snow covered roads is the depression in freezing point which caused by the addition of non-volatile impurities to a liquid. Addition of salt (sodium chloride) lowers the freezing point temperature of water and thus, helps in the melting of snow.

Question. What is “semi permeable membrane”?
Answer. The membranes which allow only the movement of the solvent molecules through them is called semi permeable membrane. The membranes appear to be continues sheet or flims. here only the molecules of the solvent can pass while those of the solute which are of bigger size, are not in a position to pass through.

Question. What type of intermolecular attractive interaction exists in the pair of methanol and acetone?
Answer. Solute-solvent dipolar interactions exist in the pair of methanol and acetone.

Question. On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
(ii) What happens when we place the blood cell in water (hypotonic solution)? Give reason.
Answer. (i) Volume decreases by mixing X and Y. It shows negative deviations from Raoult’s law. There will be rise in temperature. (ΔHmix < 0) (ii) Blood cell will swell due to osmosis as water enters the cell.

LONG ANSWER QUESTIONS

Question. a) 18 g of glucose, C₆H₁₂O₆ (Molar mass – 180 g mol^-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K)
Answer.
We know that : Elevation of boiling point ΔTb
WBMB×100×Kbwt. of solvent
Given: WB = 18 g
MB = Formula of glucose is C₆H₁₂O₆
= 6 × 12 + 12 + 6 × 16 = 180
Wt. of solvent = 1 kg or 1000 g,
Kb = 0.52 K kg mol^-1
Hence, ΔT_b = 18g180×1000×0.521000g = 0.52 K ∴B.P of the solution = 373.15 + 0.052 =373.202 K

Question. Define osmotic pressure of a solution.?How is the osmotic pressure related to the concentration of a solute in a solution?
Answer. Osmotic pressure : It is the external pressure which is applied on the side solution which is sufficient to prevent the entry of the solvent through semi-permeable membrane. According to the Boyle-van’t Hoff Law, the osmotic pressure (π) of a dilute solution is directly proportional to its molar concentration provided temperature is constant.
π ∝ C (At constant temperature)
π ∝ CT (At constant concentration)
π = CRT (R = Solution constant)

CASE BASED QUESTIONS

1. Read the following paragraph and answer the questions:

Colligative properties of a solution depend upon the number of moles of the solute dissolved and do not depend upon the nature of the solute. However, they are applicable only to dilute solutions in which the solutes do not undergo any association or dissociation. For solutes undergoing such changes, Van't Hoff introduced a factor, called Van't Hoff factor (i). This has helped not only to explain the abnormal molecular masses of such solutes in the solution but has also helped to calculate the degree of association or dissociation.

Question. What is Van’t Hoff factor (i) for a compound undergoing tertramerization in an organic solvent?
Answer. i= ¼ = 0.25

Question. Arrange the following in the increasing order of freezing point
0.1M Al₂(SO₄)₃, 0.1M KCl, 0.1M Glucose, 0.1M K₂SO₄
Answer. 0.1M Al₂(SO₄)₃, 0.1M K₂SO₄, 0.1M KCl, 0.1M Glucose

Question. The molar mass of Sodium Chloride determined by elevation of boiling point method is found to be abnormal. Why?
Answer. Elevation of boiling point is a colligative property. Since Sodium chloride dissociates in the solution we get abnormal molecular mass.

Question. What is the elevation of boiling point of a solution of 13.44g of CuCl₂ in 1kg of water?
(Kb for water = 0.52Kkg/mol^-1, molar mass of CuCl₂ = 134.4g/mol)
Answer. ΔTb = iKbm
= 3 x 0.52 x 0.1
= 1.56 K

Question. Equimolal solutions of NaCl and BaCl₂ are prepared in water. Freezing pint of NaCl is found to be -20C. What freezing point do you expect for BaCl2 solution?
Answer. 
i for NaCl = 2, i for BaCl₂ = 3
ΔTf NaCl = 2
ΔTf BaCl₂ 3
Hence Tf for BaCl₂ = -3⁰C

Question. What type of intermolecular attractive interaction exists in the pair of methanol and acetone?
Answer. - Solute-solvent dipolar interactions exist in the pair of methanol and acetone

Question. What mass of NaCl must be dissolved in 65g of water to lower the freezing point of water by 7.50^oC? The freezing point depression constant (Kf) for water is 1.86oC/m.Assume van’t Hoff factor for NaCl is 1.87(Molar mass of NaCl =58.5g)?
Answer. - ΔT=ixKfxwBx1000/m_BxW_A
ΔT=7.5, i=1.87, Kf=1.86km^-1, w_A=65g, m_B=58.5g/mol, w_B=?
Putting the values, we get,
7.5=1.87x1.86xwBx1000/58.5x65
w_B=8.2g

Question. Out of BaCl₂ and KCl, which one is more effective in causing coagulation of a negatively charged colloidal Sol? Give reason?
Answer. - BaCl₂ is more effective in causing coagulation because it has double +ve charge than K+.

Question. A 1.00 molal aqueous solution of trichloroacetic acid (CCl₃COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol^-1)
Answer. - ΔT_b= iK_bm
(100.18 – 100) °C = i × 0.512 K kg mol^-1 × 1 m
0.18 K = i × 0.512 K kg mol^-1 × 1 m
∴ i = 0.3

Question. Explain why aquatic species are more comfortable in cold water rather than in warm water?
Answer. - Aquatic species need dissolved oxygen for breathing. As solubility of gases decreases with increase of temperature, less oxygen is available in summer in the lake. Hence the aquatic species feel more comfortable in winter (low temperature) when the solubility of oxygen is higher.

Question. 18 g of glucose, C₆H₁₂O₆ (Molar mass – 180 g mol^-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K)?
Answer. - We know that: Elevation of boiling point ΔT_b,
ΔT_b=K_bxm (m-molality) Hence, ΔT_b = 18×1000×0.52/180 = 0.52 K ∴B.P of the solution = 373.15 + 0.052 = 373.202 K

Question. What is meant by ‘reverse osmosis’?
Answer. - If a pressure higher than the osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through semipermeable membrane. This process is called reverse osmosis.

Question. How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it?
Answer. - The vapour pressure of a solvent decreases when a non-volatile solute is dissolved in it because some solvent molecules are replaced by the molecules of solute

Question. At 25^oC, the vapour pressure of pure water is 23.76mm of Hg and that of an aqueous solution of urea is 22.98mm of Hg. Calculate the molality of the solution?
Answer. - We know that,
P^o-P/Po=X_B=Mole fraction of solute
23.76-22.98/23.76=X_B
X_B=0.0328
molality of the solution ‘m’ =XBx1000/ (1-0.0328) x18=1.88

Question. Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain?
Answer.- Osmosis : The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis. Osmotic pressure : The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the osmotic pressure. The osmotic pressure method has the advantage that it uses molarities instead of molalities and it can be measured at room temperature.

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?
Answer. - Henry’s law: “The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.” Applications of Henry’s law:
• In the production of carbonated beverages which are prepared under high pressure.
• Deep sea divers depend upon compressed air for their oxygen supply. (ii) Raoult’s law: For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. P = P°x Non-ideal solution shows positive and negative deviations from Raoult’s law.

Question. Define the following terms: (i) Ideal solution (ii) Azeotrope?
Answer. - Ideal solution: An ideal solution is that which obeys Raoult’s law and in which the intermolecular interactions between the different components are of same magnitude as that is found in pure components. (ii) Azeotrope: It is a type of liquid mixture having a definite composition and boiling like a pure liquid, (distils without change in compositions)