jee main chemistry important questions | Solutions jee main questions FREE

Q 1 :

We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01 M and 0.001 M, respectively. The value of van’t Hoff factor (i) for these solutions will be in the order: [2024]

$i_{A} < i_{B} < i_{C}$
$i_{A} < i_{C} < i_{B}$
$i_{A} = i_{B} = i_{C}$
$i_{A} > i_{B} > i_{C}$

1

2

3

4

(1)

Van't Hoff factor (i) is given by:

$i = \frac{Number of particles after dissociation}{Number of particles before dissociation}$

With dilution, degree of dissociation increases hence $i$ increases with dilution. Hence $i$ for A (0.1M NaCl), B (0.01M NaCl) and C (0.001M NaCl) is: $i_{A} < i_{B} < i_{C}$

Q 2 :

A solution containing 10 g of an electrolyte ${AB}_{2}$ in 100 g of water boils at 100.52 ${}^{°}C$ . The degree of ionization of the electrolyte ( $α$ ) is _____ $\times 10^{- 1}$ . (nearest integer)

[Given : Molar mass of ${AB}_{2} = 200 g$ ${mol}^{- 1}$ . $K_{b}$ (molal boiling point elevation const. of water) = 0.52 K kg ${mol}^{- 1}$ , boiling point of water = 100°C; ${AB}_{2}$ ionises as ${AB}_{2} \to A^{2 +} + 2 B^{-}$ ] [2024]

0%

(5)

$Δ T_{b} = T_{b} - T_{b}^{°} = (100.52 - 100) K = 0.52 K$

$i m k_{b} = 0.52 K$

$i \frac{w_{solute} \times 1000}{M_{solute} \times W_{solvent} (in g)} k_{b} = 0.52 K$

$i \frac{10 \times 1000}{200 \times 100} \times 0.52 = 0.52$

$i = 2$

${AB}_{2} ⇌ A^{2 +} + 2 B^{-}$

$i = 1 + (n - 1) α$

$2 = 1 + (3 - 1) α$

$α = 0.5 = 5 \times 10^{- 1}$

Q 3 :

Arrange the following solutions in order of their increasing boiling points. [2025]

(i) $10^{- 4}$ M NaCl
(ii) $10^{- 4}$ M Urea
(iii) $10^{- 3}$ M NaCl
(iv) $10^{- 2}$ M NaCl

(ii) < (i) ≡ (iii) < (iv)
(iv) < (iii) < (i) < (ii)
(ii) < (i) < (iii) < (iv)
(i) < (ii) < (iii) < (iv)

1

2

3

4

(3)

$NaCl \to {Na}^{+} + {Cl}^{-}$

$i = 2$

Urea is non-electrolyte so, $i = 1$ .

Elevation in boiling point $Δ T_{b} = i k_{b} m .$ So higher is concentration and more is $i$ , more will be the boiling point.

Q 4 :

${CrCl}_{3} \cdot {xNH}_{3}$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558 ${}^{o}C$ . Assuming 100% ionisation of this complex and coordination number of Cr is 6, the complex will be

(Given $K_{f}$ = 1.86 K kg ${mol}^{- 1}$ ) [2025]

$[Cr {({NH}_{3})}_{5} Cl] {Cl}_{2}$
$[Cr {({NH}_{3})}_{6}] {Cl}_{3}$
$[Cr {({NH}_{3})}_{3} {Cl}_{3}]$
$[Cr {({NH}_{3})}_{4} {Cl}_{2}] Cl$

1

2

3

4

(1)

$Δ T_{f} = i k_{f} m$

$0.558 = i \times 1.86 \times 0.1$

$i = 3$

$[Cr {({NH}_{3})}_{5} Cl] {Cl}_{2} \to {[Cr {({NH}_{3})}_{5} Cl]}^{2 +} + 2 {Cl}^{-},$ $n = 3$

Q 5 :

1.24 g of $A X_{2}$ (molar mass 124 g ${mol}^{- 1}$ ) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156°C, while 25.4 g of $A Y_{2}$ (molar mass 250 g ${mol}^{- 1}$ ) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.

$K_{b} (H_{2} O) = {0.52 K kg mol}^{- 1}$

Which of the following is correct? [2025]

${AX}_{2}$ and $A Y_{2}$ (both) are completely unionised.
${AX}_{2}$ is fully ionised while $A Y_{2}$ is completely unionised.
${AX}_{2}$ and $A Y_{2}$ (both) are fully ionised.
${AX}_{2}$ is completely unionised while $A Y_{2}$ is fully ionised.

1

2

3

4

(2)

$Moles of A X_{2} (n_{A X_{2}}) = \frac{W_{A X_{2}}}{M_{A X_{2}}} = \frac{1.24}{124} mol = 0.01 mol$

$Molality of A X_{2} solution (m_{A X_{2}}) = \frac{n_{A X_{2}}}{W_{water (in kg)}} = \frac{0.01 mol}{1 kg} = 0.01 m$

$Elevation in boiling point of A X_{2} solution (Δ T_{b} (A X_{2})) = (100.0156 - 100) ° C = i_{A X_{2}} m_{A X_{2}} k_{b}$

$0.0156 = i_{A X_{2}} \times 0.01 \times 0.52$

$i_{A X_{2}} = \frac{0.0156}{0.01 \times 0.52} = 3$

So, $A X_{2}$ is completely ionized.

$Moles of A Y_{2} (n_{A Y_{2}}) = \frac{W_{A Y_{2}}}{M_{A Y_{2}}} = \frac{25.4}{250} mol = 0.1016 mol$

$Molality of A Y_{2} solution (m_{A Y_{2}}) = \frac{n_{A Y_{2}}}{W_{water (in kg)}} = \frac{0.1016 mol}{2 kg} = 0.0508 m$

$Elevation in boiling point of A Y_{2} solution (Δ T_{b} (A Y_{2})) = (100.026 - 100) ° C = i_{A Y_{2}} m_{A Y_{2}} k_{b}$

$0.026 = i_{A Y_{2}} \times 0.0508 \times 0.52$

$i_{A Y_{2}} = \frac{0.026}{0.0508 \times 0.52} \approx 1$

So, $A Y_{2}$ is completely unionized.

Q 6 :

The observed and normal molar masses of compound ${MX}_{2}$ are 65.6 and 164 respectively. The percent degree of ionisation of ${MX}_{2}$ is _____ %. (Neatest integer) [2025]

0%

(75)

$i = \frac{normal molar mass}{observed molar mass} = \frac{164}{65.6} = 2.5$

$MX ⇌ M^{2 +} + 2 X^{-}$

Moles in which one mole of $M X_{2}$ dissociates $(n) = 3$

$1 + (n - 1) α$ = 2.5

$1 + (3 - 1) α$ = 2.5

$α$ = 0.75 = 75%

Q 7 :

If $A_{2} B$ is 30% ionised in an aqueous solution, then the value of van’t Hoff factor $(i)$ is _____ $\times 10^{- 1}$ . [2025]

0%

(16)

$A_{2} B ⇌ 2 A^{+} + B^{2 -}$

$i = 1 + (n - 1) α$

One molecule of $A_{2} B$ dissociates into three particles, so $n$ = 3.

$α$ = 0.3 as per question.

$i = 1 + (3 - 1) \times 0.3$

$i = 1.6 = 16 \times 10^{- 1}$

Q 8 :

The percentage dissociation of a salt ( ${MX}_{3}$ ) solution at given temperature (van’t Hoff factor $i$ = 2) is ................. % (Nearest integer) [2025]

0%

(33)

${MX}_{3} \to M^{3 +} + 3 X^{-}$

Van’t Hoff factor $(i)$ is related to degree of dissociation $(α)$ as:

$i = 1 + (n - 1) α$

Where $n$ is the number of moles formed by complete dissociation of 1 mol of salt

$2 = 1 + (3 - 1) α$

$α$ = 0.3333

Percentage dissociation = 33.33% (nearest integer = 33)

Q 9 :

If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be _______% higher than the expected/theoretical freezing point. (Nearest integer) [2023]

0%

(30)

For monobasic acid, $n = 2$

$i = 1 + (n - 1) α = 1 + (2 - 1) \times 0.3 = 1.3$

$% increase = \frac{{(Δ T_{f})}_{_{obs}} - {(Δ T_{f})}_{cal}}{{(Δ T_{f})}_{cal}} \times 100$

$= \frac{i \times K_{f} \times m - K_{f} \times m}{K_{f} \times m} \times 100$

$= (i - 1) \times 100 = 0.3 \times 100 = 30 %$

Q 10 :

0.004 M $K_{2} {SO}_{4}$ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of $K_{2} {SO}_{4}$ is _______% (Nearest integer) [2023]

0%

(75)

$Isotonic solution,$

$π_{K_{2} S O_{4}} = π_{Glucose}$

$i \times 0.004 \times R T = 0.01 \times R T$

$i = 2.5$

For $K_{2} S O_{4}$ {for dissociation $i = 1 + (n - 1) α$ }

$DOD (α) = \frac{i - 1}{n - 1} = \frac{2.5 - 1}{3 - 1} = 0.75$

$% dissociation = 75$

Topic Question Set

Q 1 :

We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01 M and 0.001 M, respectively. The value of van’t Hoff factor (i) for these solutions will be in the order: [2024]

0%

Q 3 :

Arrange the following solutions in order of their increasing boiling points. [2025]

(i) $10^{- 4}$ M NaCl
(ii) $10^{- 4}$ M Urea
(iii) $10^{- 3}$ M NaCl
(iv) $10^{- 2}$ M NaCl

Q 6 :

The observed and normal molar masses of compound ${MX}_{2}$ are 65.6 and 164 respectively. The percent degree of ionisation of ${MX}_{2}$ is _____ %. (Neatest integer) [2025]

0%

Q 7 :

If $A_{2} B$ is 30% ionised in an aqueous solution, then the value of van’t Hoff factor $(i)$ is _____ $\times 10^{- 1}$ . [2025]

0%

Q 8 :

The percentage dissociation of a salt ( ${MX}_{3}$ ) solution at given temperature (van’t Hoff factor $i$ = 2) is ................. % (Nearest integer) [2025]

0%

Q 9 :

If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be _______% higher than the expected/theoretical freezing point. (Nearest integer) [2023]

0%

Q 10 :

0.004 M $K_{2} {SO}_{4}$ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of $K_{2} {SO}_{4}$ is _______% (Nearest integer) [2023]

0%

Want Us to Email you About Special Offers & Updates?