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Q 11 :
The solution from the following with highest depression in freezing point/lowest freezing point is: [2024]

180 g of acetic acid dissolved in water
180 g of acetic acid dissolved in benzene
180 g of benzoic acid dissolved in benzene
180 g of glucose dissolved in water

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(1)

Number of moles of acetic acid:

$(n_{C H_{3} C O O H}) = \frac{Given mass}{Molar mass} = \frac{180 g}{60 g m o l^{- 1}} = 3 m o l$

Number of moles of benzoic acid:

$(n_{C_{6} H_{5} C O O H}) = \frac{Given mass}{Molar mass} = \frac{180 g}{122 g m o l^{- 1}} = 1.475 m o l$

Number of moles of glucose:

$(n_{C_{6} H_{12} O_{6}}) = \frac{Given mass}{Molar mass} = \frac{180 g}{180 g m o l^{- 1}} = 1 m o l$

In water, carboxylic acids undergo dissociation, resulting in an increase in the number of particles, consequently contributing towards increasing the value of the colligative property.

In benzene, carboxylic acids undergo dimerisation, resulting in a decrease in the number of particles, consequently contributing towards decreasing the value of the colligative property.

Glucose does not dissociate or associate in water.
As the most number of particles are produced in (1), it has the highest depression in freezing point.
Assuming complete dissociation of acetic acid in water, $i = 2$ .
Assuming complete association of acetic acid and benzoic acid in benzene $(i = \frac{1}{2})$ .
Assuming 1 kg of solvent in each case, various $Δ T_{f}$ values are as follows:

Solutions	Depression in freezing point	Freezing point of solution
180 g acetic acid in 1 kg water	or (change in temperature has same value in K or ^oC)
180 g acetic acid in 1 kg benzene	or
180 g benzoic acid in 1 kg benzene	or
180 g glucose in 1 kg water	or

Q 12 :
Consider the given plots of vapour pressure (VP) vs temperature (T/K). Which amongst the following options is correct graphical representation showing $∆ T_{f},$ depression in the freezing point of a solvent in a solution? [2025]

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(3)

Q 13 :
What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g ${mol}^{- 1}$ ) and the decrease in freezing point is 0.40 K? [2025]

5.12 K kg ${mol}^{- 1}$
3.72 K kg ${mol}^{- 1}$
4.43 K kg ${mol}^{- 1}$
1.86 K kg ${mol}^{- 1}$

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(1)

$Δ T_{f} = m k_{f}$

$Δ T_{f} = \frac{w_{solute} (in g) \times 1000}{M_{solute} W_{solvent} (in g)} k_{f}$

$0.4 = \frac{1 \times 1000}{256 \times 50} k_{f}$

$k_{f} = \frac{0.4 \times 256 \times 50}{1000} {K kg mol}^{- 1}$

$= 5.12 {K kg mol}^{- 1}$

Q 14 :
Given below are two statements: [2025]

Statement I: NaCl is added to the ice at 0°C, present in the ice cream box to prevent the melting of ice cream.

Statement II: On addition of NaCl to ice at 0°C, there is a depression in freezing point.

In the light of the above statements, choose the correct answer from the options given below:

Both Statement I and Statement II are false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Statement I is true but Statement II is false

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(2)

When a non volatile solute is added to a solvent, its freezing point decreases.

Q 15 :
2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is:

(Given: Ebulioscopic constant of water = 0.52 K kg ${mol}^{- 1}$ ) [2025]

379.2 K
377.3 K
375.3 K
277.3 K

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(2)

Total moles of solute $= 2 mol glycol+ 2 mol glucose = 4 mol$

Molality of solution $(m)$

$= \frac{Total moles of electrolyte}{Mass of solvent in gm} \times 1000$

$= \frac{4}{500} \times 1000 = 8 m$

Elevation in boiling point, $Δ T_{b} = m K_{b} = 8 \times 0.52 K = 4.16 K$

Boiling point of solution $(T_{b} (solution))$

$= T_{b} (water) + Δ T_{b} = (373.15 + 4.16) K = 377.31 K$

Q 16 :
XY is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration $c_{1}$ and $c_{2}$ $(c_{1} > c_{2})$ mol $L^{- 1}$ . For the reverse osmosis to take place identify the correct condition

(Here $p_{1}$ and $p_{2}$ are pressures applied on chamber 1 and 2)

(A) Membrane/Partition ; Cellophane, $p_{1} > π$

(B) Membrane/Partition ; Porous, $p_{2} > π$

(C) Membrane/Partition ; Parchment paper, $p_{1} > π$

(D) Membrane/Partition ; Cellophane, $p_{2} > π$

Choose the correct answer from the option given below :

B and D only
A and D only
A and C only
C only

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(3)

During osmosis, solvent moves from low concentration solution to high concentration solution, across a semipermeable membrane. For reverse osmosis, we need to apply a pressure greater than osmotic pressure, on solution of high concentration, so that solvent moves against natural flow of osmosis i.e. from high concentration $(c_{1})$ to low concentration $(c_{2})$ .

Q 17 :
Given below are two statements : [2025]

Statement (I): Molal depression constant $K_{f}$ is given by $\frac{M_{1} R T_{f}}{Δ S_{fus}},$ where symbols have their usual meaning.

Statement (II): $K_{f}$ for benzene is less than the $K_{f}$ for water.

In the light of the above statements, choose the most appropriate answer from the options given below:

Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Both Statement I and Statement II are correct
Statement I is correct but Statement II is incorrect

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(4)

Statement I: Molal depression constant $(k_{f})$ = $= \frac{M_{1} R {(T_{f} °)}^{2}}{Δ H_{fus}},$ where $M_{1}$ is molar mass of solvent, R is gas constant, $T_{f} °$ is freezing point of solvent, $Δ H_{fus}$ is molar enthalpy of fusion of solvent.

$k_{f} = \frac{M_{1} R {(T_{f} °)}^{2}}{Δ H_{fus}} = \frac{M_{1} R T_{f} °}{Δ H_{fus} / T_{f} °} = \frac{M_{1} R T_{f} °}{Δ S_{fus}}$

Statement II:

$k_{f} (benzene) = 5.12 K kg/mol$

$k_{f} (water) = 1.86 K kg/mol$

Q 18 :
$HA(aq) ⇌ H^{+} (a q) + A^{-} (a q)$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is $0.20 ° C .$

The dissociation constant for the acid is

Given: $K_{f} (H_{2} O) = 1.8 {K kg mol}^{- 1}, molality \equiv molarity$ [2025]

$1.38 \times 10^{- 3}$
$1.1 \times 10^{- 2}$
$1.90 \times 10^{- 3}$
$1.89 \times 10^{- 1}$

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(1)

$Δ T_{f} = k_{f} m$

$0.2 = i \times 1.8 \times 0.1$

$i = \frac{0.2}{1.8 \times 0.1} = \frac{10}{9}$

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

$i = 1 + (n - 1) α$

Where $n$ is number of moles in which 1 mol of HA completely dissociates.

$\frac{10}{9} = 1 + (2 - 1) α$

$α = \frac{1}{9}$

$K_{a} = \frac{C α^{2}}{1 - α} = \frac{0.1 {(\frac{1}{9})}^{2}}{1 - \frac{1}{9}} = 1.38 \times 10^{- 3}$

Q 19 :
When 1 g each of compounds AB and ${AB}_{2}$ are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu) is ______ $\times 10^{- 1}$ (Nearest integer).

(Given: Molal boiling point elevation constant is 0.5 K kg ${mol}^{- 1}$ ) [2025]

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(25)

Solution of AB:

$Δ T_{b} = m_{A B} K_{b}$

$Δ T_{b} = \frac{W_{A B}}{M_{A B}} \times \frac{1000}{W_{water}} K_{b}$

$2.7 = \frac{1 \times 1000}{M_{A B} \times 15} \times 0.5$

$M_{A B} = 12.346$

Thus, $M_{A} + M_{B} = 12.346$ ...(I)

Solution of ${AB}_{2}$ :

$Δ T_{b} = m_{A B_{2}} K_{b}$

$Δ T_{b} = \frac{W_{A B_{2}}}{M_{A B_{2}}} \times \frac{1000}{W_{water}} K_{b}$

$1.5 = \frac{1 \times 1000}{M_{A B_{2}} \times 15} \times 0.5$

$M_{A B_{2}} = 22.22 g$

Thus, $M_{A} + 2 M_{B} = 22.22$ ...(II)

(II) – (I) gives us:

$M_{B} = 22.22 - 12.346 = 9.874 g$

Putting this in (I) gives us:

$M_{A} + 9.874 = 12.346$

$M_{A} = 2.472 g = 25 \times 10^{- 1}$

Topic Question Set

Q 11 :
The solution from the following with highest depression in freezing point/lowest freezing point is: [2024]

Q 12 :
Consider the given plots of vapour pressure (VP) vs temperature (T/K). Which amongst the following options is correct graphical representation showing $∆ T_{f},$ depression in the freezing point of a solvent in a solution? [2025]

(3)

Q 13 :
What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g ${mol}^{- 1}$ ) and the decrease in freezing point is 0.40 K? [2025]

Q 15 :
2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is:

(Given: Ebulioscopic constant of water = 0.52 K kg ${mol}^{- 1}$ ) [2025]

Q 18 :
$HA(aq) ⇌ H^{+} (a q) + A^{-} (a q)$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is $0.20 ° C .$

The dissociation constant for the acid is

Given: $K_{f} (H_{2} O) = 1.8 {K kg mol}^{- 1}, molality \equiv molarity$ [2025]

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Topic Question Set

Q 11 : The solution from the following with highest depression in freezing point/lowest freezing point is: [2024]

Q 12 : Consider the given plots of vapour pressure (VP) vs temperature (T/K). Which amongst the following options is correct graphical representation showing ∆Tf,depression in the freezing point of a solvent in a solution? [2025]

(3)

Q 13 : What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g mol-1) and the decrease in freezing point is 0.40 K? [2025]

Q 15 : 2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given: Ebulioscopic constant of water = 0.52 K kg mol-1) [2025]

Q 18 : HA(aq)⇌H+(aq)+A-(aq) The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20°C. The dissociation constant for the acid is Given: Kf(H2O)=1.8 K kg mol-1, molality≡molarity [2025]

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Q 11 :
The solution from the following with highest depression in freezing point/lowest freezing point is: [2024]

Q 12 :
Consider the given plots of vapour pressure (VP) vs temperature (T/K). Which amongst the following options is correct graphical representation showing $∆ T_{f},$ depression in the freezing point of a solvent in a solution? [2025]

Q 13 :
What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g ${mol}^{- 1}$ ) and the decrease in freezing point is 0.40 K? [2025]

Q 15 :
2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is:

(Given: Ebulioscopic constant of water = 0.52 K kg ${mol}^{- 1}$ ) [2025]

Q 18 :
$HA(aq) ⇌ H^{+} (a q) + A^{-} (a q)$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is $0.20 ° C .$

The dissociation constant for the acid is

Given: $K_{f} (H_{2} O) = 1.8 {K kg mol}^{- 1}, molality \equiv molarity$ [2025]