Students can read the important questions given below for **Solutions Class 12** Chemistry. All Solutions Class 12 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 12 Chemistry Important Questions provid ed for all chapters to get better marks in examinations. Chemistry Question Bank Class 12 is available on our website for free download in PDF.

**Important Questions of Solutions Class 12**

### Very Short Answer Questions

**Question. **What is meant by molality of a solution?**Answer. **Molality of a solution can be defined as the number of moles of solute dissolved in one kg solvent. It is denoted by m.

**Question. **State the main advantage of molality over molarity as the unit of concentration.**Answer. **Molality is independent of temperature, whereas molarity is a function of temperature.

**Question. **Gas (A) is more soluble in water than gas (B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry’s constant) and why?**Answer. ** According to Henry’s law, the solubility of a gas is inversely proportional to the Henry’s law constant (K_{H}) for that gas. Hence, gas (B) being less soluble, would have a higher K_{H} value.

**Question. **Explain the following :**Henry’s law about dissolution of a gas in a liquid.****Answer. **Henry’s law states that, the partial pressure of the gas in vapour phase (p) is proportional to the

mole fraction of the gas (x) in the solution.

p = K_{H}⋅x where, K_{H} = Henry’s law constant.

Different gases have different K_{H} values at the same temperature.

**Question. **Some liquids on mixing form ‘azeotropes’.**What are ‘azeotropes’?****Answer. **Azeotropes are the binary mixtures of solutions that have the same composition in liquid and vapour phases and that have constant boiling points. It is not possible to separate the components of azeotropes by fractional distillation.

**Question. **Defne the following term :**Ideal solution****Answer. ** A solution which obeys Raoult’s law at all concentrations and temperatures is called an ideal solution.

**Question. **How is it that alcohol and water are miscible in all proportions**Answer. **Both alcohol and water are polar in nature hence, they are miscible in all proportions. Water and ethanol molecules attract

each other because of the formation of H-bonds. This property also makes them miscible.

**Question. **State the condition resulting in reverse osmosis.**Answer. **Reverse osmosis occurs when a pressure larger than the osmotic pressure is applied to the solution side.

**Question.** Defne the following term :**Van’t Hoff factor****Answer. **van’t How factor : It is defined as the ratio of the experimental value of colligative property to the calculated value of the colligative property and is used to find out the extent of dissociation or association. Mathematically, it is represented as

### Short Answer Questions

**Question. **An antifreeze solution is prepared from 222.6 g of ethylene glycol (C_{2}H_{4}(OH)_{2}) and 200 g of water. Calculate the molality of the solution.**If the density of this solution be 1.072 g mL ^{–1} what will be the molality of the solution?**

**Answer.**Mass of the solute, C

_{2}H

_{4}(OH)

_{2}= 222.6 g

Molar mass of solute, C

_{2}H

_{4}(OH)

_{2}= 62 g mol

^{–1}

**Question. **(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. **(i) The elevation in boiling point of a solution is a colligative property which depends on the number of moles of solute added. Higher the concentration of solute added, higher will be the elevation in boiling point. Thus, 2 M glucose has higher boiling point than 1 M glucose solution.

(ii) When the external pressure applied becomes more than the osmotic pressure of solution then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side and the process is called reverse osmosis.

**Question. **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. **(i) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells thus, on placing blood cells in this solution exosmosis takes place that results in shrinking of cells.

(b) 0.4% sodium chloride solution is hypotonic with respect to 0.9% sodium chloride solution or blood cells thus, on placing blood cells in this solution endosmosis takes place that results in swelling of cells.

**Question. **Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?**Answer. **The boiling point of the solution is always higher than that of the pure solvent. As the vapour pressure of the solution is lower than that of the pure solvent and vapour pressure increases with increase in temperature. Hence, the solution has to be heated more to make the vapour pressure equal to the atmospheric pressure. Elevation of boiling point is a colligative property because it depends on number of solute particles present in a solution.

**Question. **Calculate the mass of compound (molar mass = 256 g mol^{–1}) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K.**(Kf = 5.12 K kg mol ^{–1}).**

**Answer.**Given : W

_{2}= ?, M

_{2}= 256 g mol

^{–1}, ΔT

_{f}= 0.48 K

W

_{1}= 75 g, K

_{f}= 5.12 K kg mol

^{–1}

**Question. **18 g of glucose, C_{6}H_{12}O_{6} (Molar mass = 180 g mol^{–1}) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil?**( K_{b} for water = 0.52 K kg mol^{–1}, boiling point of pure water = 373. 15 K)**

**Answer.**Given W

_{1}= 1 kg = 1000 g, W

_{2}= 18 g,

M

_{2}= 180 g mol

^{–1}

T°

_{b}= 373.15 K, K

_{b}= 0.52 K Kg mol

^{–1}, T

_{b}= ?

Using formula,

**Question. **Outer hard shells of two eggs are removed.**One of the egg is placed in pure water and the other is placed in saturated solution of sodium chloride. What will be observed and why?****Answer. **The egg placed in pure water will swell because the concentration of proteins is high inside the egg as compared to water. Therefore, endosmosis occurs and water diffuses through the semipermeable membrane. The egg which is placed in sodium chloride solution will shrink due to osmosis of water out of the egg.

**Question. **Find the boiling point of a solution containing 0.520 g of glucose (C_{6}H_{1}2O_{6}) dissolved in 80.2 g of water.**[Given : K_{b} for water = 0.52 K/m]**

**Answer.**Given, W

_{2}= 0.520 g,

W

_{1}= 80.2 g, Kb = 0.52 K m

^{–1}

M

_{2}of C

_{6}H

_{12}O

_{6}= 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol

^{–1}

### Long Answer Questions

**Question. **Calculate the freezing point of solution when 1.9 g of MgCl_{2}(M = 95 g mo**l ^{–1}**) was dissolved in 50 g of water, assuming MgCl

_{2}undergoes complete ionization.

**(Kf for water = 1.86 K kg mo**

**l**)^{–1}**Answer.**

**Question. **When 2.56 g of sulphur was dissolved in 100 g of CS_{2}, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).**(K _{f} the CS_{2} = 3.83 K kg mol^{–1}, atomic mass of sulphur = 32 g mol^{–1})**

**Answer.**W

_{2}= 2.56 g, W

_{1}= 100 g, ΔT

_{f}= 0.383 K

K

_{f}= 3.83 K kg mol

^{–1}, ΔT

_{f}= K

_{f }× m

**Question. **Calculate the boiling point of solution when 4 g of MgSO_{4} (M = 120 g mo**l ^{–1}**) was dissolved in 100 g of water, assuming MgSO

_{4}undergoes complete ionization.

**(K**

_{b}for water = 0.52 K kg mo**l**)^{–1}**Answer.**W

_{2}= 4 g, M

_{2 }= 120 g mol

^{–1}

W

_{1}= 100 g, K

_{b}= 0.52 K kg mol

^{–1}

For complete dissociation, i = 2

Using formula, ΔTb = iK

_{b}m

**Question. **Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K (the K_{f} for water = 1.86 K kg mo^{l–1}). **Answer. **ΔT_{f }= 2 K, K_{f} = 1.86 K kg mol^{–1},

W_{1} = 1 kg, ΔT_{f} = i K_{f} m, M_{2}(KCl) = 74.5 g mol^{–1}

i = 2 for KCl

**Question. **Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr_{2} in 200 g of water. (Molar mass of MgBr_{2} = 184 g mo**l ^{–1}**)

**(K**

_{f }for water = 1.86 K kg mo**l**)^{–1}**Answer.**

W_{2 }= 10.50 g, W_{1} = 200 g

M_{2}(MgBr_{2}) = 184 g mol^{–1}

K_{f }= 1.86 K kg mol^{–1}

**Question. **A 0.561 m solution of an unknown electrolyte depresses the freezing point of water by 2.93°C. What is van’t Hoff factor for this electrolyte?**The freezing point depression constant (K _{f}) for water is 1.86°C kg mol^{–1}.**

**Answer.**m = 0.561 m, ΔT

_{f }= 2.93°C and

K

_{f}= 1.86°C kg mol

^{–1}

ΔT

_{f}= iK

_{f}m

**Question. **Phenol associates in benzene to a certain extent to form a dimer. A solution containing 20 g of phenol in 1.0 kg of benzene has its freezing point lowered by 0.69 K. Calculate the fraction of phenol that has dimerised **[Given K _{f} for benzene = 5.1 K m^{–1}]**

**Answer.**Here, n = 2 because phenol forms dimer on association.

W

_{2}= 20 g, W

_{1 }= 1 kg = 1000 g, ΔT

_{f }= 0.69 K,

K

_{f}= 5.1 K m

^{–1}

**Question. **The boiling point elevation of 0.30 g acetic acid in 100 g benzene is 0.0633 K. Calculate the molar mass of acetic acid from this data. What conclusion can you draw about the molecular state of the solute in the solution?**[Given K _{b} for benzene = 2.53 K kg mol^{–1}]**

**Answer.**

Here, i < 1, therefore the solute acetic acid is associated in benzene

**Question. **The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by 0.45°C. Calculate.**(i) the molar mass of acetic acid from this data****(ii) van’t Hoff factor****[For benzene, K _{f} = 5.12 K kg mol^{–1}] What conclusion can you draw from the value of van’t Hoff factor obtained?**

**Answer.**

**Question. **Calculate the freezing point of the solution when 31 g of ethylene glycol (C_{2}H_{6}O_{2}) is dissolved in 500 g of water.**(K _{f} for water = 1.86 K kg mol^{–1})**

**Answer.**Mass of ethylene glycol (C

_{2}H

_{6}O

_{2}), W

_{2}= 31 g

Mass of water, W1 = 500 g

M

_{2 }(Mol. mass of C

_{2}H

_{6}O

_{2}) = 62 g mol

^{–1},

K

_{f}= 1.86 K kg mol

^{–1}, T

_{f}= ?

**Question. **1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mo**l ^{–1}**. Find the molar mass of the solute.

**Answer.**W

_{2}= 1.00 g, W

_{1}= 50 g, K

_{f}= 5.12 K kg mol

^{–1},

ΔT

_{f}= 0.40 K

**Question. **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 C _{6}H_{12}O_{6} : 180 amu;**

**Cane-sugar C**

_{12}H_{22}O_{11}: 342 amu]**Answer.**Molality of sugar solution

**Question. **A solution of glycerol (C_{3}H_{8}O_{3}) 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?**(K _{b} for water = 0.512 K kg mol^{–1})**

**Answer.**W

_{1}= 500 g

Boiling point of solution (T

_{b}) = 100.42°C

K

_{b}for water = 0.512 K kg mol

^{–1}

M

_{2}(C

_{3}H

_{8}O

_{3}) = (3 × 12) + (8 × 1) + (3 × 16)

= 92 g mol

^{–1}

ΔTb = Tb – T°

_{b}

= 373.42 K – 373 K = 0.42 K

**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?**(K _{f} for water = 1.86 K kg mol^{–1})**

**Answer.**W

_{1}= 450 g, W

_{2}= 15.0 g

ΔT

_{f}= T°

_{f}– T

_{f }= 273 K – 272.66 K = 0.34 K

K

_{f}= 1.86 K kg mol

^{–1}

**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. **The relative lowering of vapour pressure is given by the following expression,

(p°solvent – psolution)/p°solvent = n_{2}/(n_{1} + n_{2})

for dilute solutions, n_{2} << n_{1}, therefore

(p°solvent – psolution)/p°solvent = n_{2}/n_{1}

= (W_{2} × M_{1})/(M_{2} × W_{1})

(p°solvent – 2.8)/p°solvent = (30 × 18)/(M_{2} × 90)

(p°solvent – 2.8)/p°solvent = 6/M_{2} …(1)

Similarly for second case we get,

(p°solvent – 2.9)/p°solvent = (30 × 18)/(M_{2} × 108)

(p°solvent – 2.9)/p°solvent = 5/M_{2} …(2)

On solving eq. (1) and (2), we get

(p°solvent – 2.8)/(p°solvent – 2.9) = 6/5**∴** p°solvent = 3.4 kPa

i.e., vapour pressure of water at 298 K is 3.4 kpa Substituting the value of p°solvent in (1) we get,

(3.4 – 2.8)/3.4 = 6/M_{2}

or 0.6/3.4 = 6/M_{2} **∴** M_{2} = 34 g

**Question. **Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.00 g of water. (K_{b} for water = 0.512 K kg mo**l ^{–1}**),

**(Molar mass of NaCl = 58.44 g)**

**Answer.**i = 2, K

_{b}= 0.512 K kg mol

^{–1}, WB = 15 g

M

_{B}= 58.44 g mol

^{–1,}W

_{A}= 250 g

Therefore, boiling point of aqueous solution,

T_{b} = T°_{b} + ΔT_{b} = 373.15 K + 1.05 K = 374.20 K

**Question. **Calculate the boiling point of one molar aqueous solution (density 1.06 g m**l ^{–1}**) of KBr.

**[Given : K**

_{b}for H_{2}O = 0.52 K kg mo**l**, atomic^{–1}**mass : K = 39, Br = 80]**

**Answer.**Concentration of the solution = 1 molar

Density of the solution = 1.06 g mL

^{–1}

M

_{2}, molar mass of KBr = 39 + 80 = 119 g mol

^{–1}

K

_{b}for H

_{2}O = 0.52 K kg mol

^{–1}

**Question. **A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound.**(B.pt. of pure benzene = 80.10°C and K _{b} for benzene = 2.53°C kg mol^{–1})**

**Answer.**W

_{2}= 1.25 g, W

_{1}= 99.0 g

ΔT

_{b}= T

_{b}– T°b = (80.31 – 80.10)°C = 0.21°C = 0.21 K

ΔT

_{b}= K

_{b}⋅m

**Question. **What mass of ethylene glycol (molar mass = 62.0 g mo**l ^{–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 mo**

**l**)^{–1}**Answer.**M

_{2}(ethylene glycol) = 62 g mol

^{–1}

W

_{1}= 5.50 kg = 5500 g, ΔT

_{f}= T°

_{f }– T

_{f }= 0°C – (–10°C)

= 10°C = 10 K and K

_{f}= 1.86 K kg mol

^{–1}

**Question. **Calculate the freezing point of a solution containing 18 g glucose, C_{6}H_{12}O_{6} and 68.4 g sucrose, C_{12}H_{22}O_{11} in 200 g of water. The freezing point of pure water is 273 K and Kf for water is 1.86 K m^{–1}.**Answer. **Molar mass of glucose, C_{6}H_{12}O_{6}

= 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^{–1}

Molar mass of sucrose, C_{12}H_{22}O_{11}

= 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol^{–1}

**Question. **Calculate the temperature at which a solution containing 54 g of glucose, (C_{6}H_{12}O_{6}), in 250 g of water will freeze.**(K _{f }for water = 1.86 K mol^{–1} kg)**

**Answer.**M

_{2}(glucose, C

_{6}H

_{12}O

_{6}) = 180 g mol

^{–1}

W

_{2}= 54 g, W

_{1}= 250 g, K

_{f}= 1.86 K kg mol

^{–1}

**Question. **A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86°C, whereas pure ether boils at 35.60°C. Determine the molecular mass of the solute. (For ether K_{b} = 2.02 K kg mo**l ^{–1}**)

**Answer.**T

_{b}= 36.86°C, T°

_{b}= 35.60°C

ΔT

_{b}= T

_{b}– T°

_{b}= 36.86 – 35.60 = 1.26°C

**Question. **A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mo**l ^{–1}**) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. mass = 180 g mol

^{–1}) per 100 g of solution.

**Answer.**Molality = 0.1539 m,

ΔT

_{f}= T°

_{f}– T

_{f}= 273.15 – 271 = 2.15 K

Again mass of solute, W_{2} = 5 g

Molar mass of solute, M_{2} = 180 g mol^{–1}

Mass of solution = 100 g**∴** Mass of solvent, W_{1} = 95 g

**Question. **The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.**Answer. **Given : p°_{A} = 450 mm Hg, p°_{B} = 700 mm Hg,

PTotal = 600 mm Hg, x_{A} = ?

Applying Raoult’s law, p_{A} = x_{A} × p°_{A}

p_{B} = x_{B} × p°_{B} = (1 – xA)p°_{B}

P_{Total} = p_{A} + p_{B} = x_{A} × p°A + (1 – x_{A})p°_{B}

= p°_{B }+ (p°_{A} – p°_{B})x_{A}

Substituting the given values, we get

600 = 700 + (450 – 700)x_{A} or, 250x_{A} = 100

**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. **Applying the relationship,

m = K_{H }× p

In the first case, 6.56 × 10^{–2} g = K_{H} × 1 bar

or, K_{H} = 6.56 × 10^{–2} g bar^{–1 }

In the second case,

5.0 × 10^{–2} g = (6.56 × 10^{–2} g bar–1) × p

**Question. **If **N _{2}** gas is bubbled through water at 293 K, how many millimoles of

**N**gas would dissolve in 1 litre of water? Assume that N

_{2}_{2}exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for

**N**at 293 K is 76.48 k bar.

_{2}**Answer.**According to Henry’s law, P

_{N2}= K

_{H}× X

_{N2}

Thus, composition of the liquid mixture will be

x_{A} = 0.40

x_{B} = 1 – 0.40 = 0.60

Calculation of composition in the vapour phase,

p_{A} = x_{A} × p°_{A} = 0.40 × 450 mm Hg = 180 mm Hg

p_{B} = x_{B} × p°_{B} = 0.60 × 700 mm Hg = 420 mm Hg

Mole fraction of A in the vapour phase