jee main chemistry important questions | Solutions jee main questions FREE

Q 1 :

What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Increases
Remains unchanged
First decreases and then increases
Decreases

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(4)

When a small quantity of naphthalene is added to benzene its freezing point decreases.

Q 2 :

The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

(73)

$Osmotic pressure(π) (for a non - electrolyte solution) is related to concentration (C) and temperature (T) as,$

$π = C R T, where R is universal gas constant$

$π_{1} = C R T_{1}, π_{2} = C R T_{2}$

$\frac{π_{2}}{π_{1}} = \frac{T_{2}}{T_{1}}$

$\frac{π_{2}}{7 \times 10^{5} Pa} = \frac{283 K}{273 K}$

$π_{2} = 7 \times 10^{5} \times \frac{283}{273} Pa$

$= 7.256 \times 10^{5} {Nm}^{- 2}$

$= 72.56 \times 10^{4} {Nm}^{- 2}$

Q 3 :

Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

(Molar mass in g ${mol}^{- 1}$ for ethylene glycol 62, $K_{f}$ of water = 1.86 K kg ${mol}^{- 1}$ ) [2024]

(15)

$Freezing point of pure water T_{f}^{o} = 0 ° C$

$Freezing point of ethylene glycol solution T_{f} = - 24 ° C$

$Depression in freezing point, Δ T_{f} = T_{f}^{o} - T_{f} = 0 ° C - (- 24 ° C) = 24 ° C$

$Value of difference of two temperatures is same in ° C and K, thus$

$Δ T_{f} = 24 ° C or 24 K$

$Δ T_{f} = i \times m \times k_{f}$

$Assuming glycol to be undissociated in water, i = 1$

$24 = 1 \times m \times 1.86$

$m = \frac{24}{1.86}$

$\frac{W_{glycol}}{M_{glycol} \times W_{water} (in kg)} = \frac{24}{1.86}$

$\frac{W_{glycol}}{62 \times 18.6} = \frac{24}{1.86}$

$W_{glycol} = 14880 g = 14.88 kg$

Q 4 :

2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°C. Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is ____ mm of Hg (nearest integer).

[Given: Molal boiling point elevation constant of water $(K_{b}) = 0.52 K . kg {mol}^{- 1}$ ,1 atm pressure = 760 mm of Hg, Molar mass of water = 18 g ${mol}^{- 1}$ ] [2024]

(711)

$Δ T_{b} = m \times k_{b}$

$2 = m \times 0.52$

$m = \frac{2}{0.52} m$

$\frac{x_{solute} \times 1000}{x_{solvent} \times M_{solvent}} = \frac{2}{0.52}$

$\frac{x_{solute} \times 1000}{x_{solvent} \times 18} = \frac{2}{0.52}$

$\frac{x_{solute}}{x_{solvent}} = \frac{2}{0.52} \times \frac{18}{1000}$

$\frac{x_{solvent}}{x_{solute}} = \frac{130}{9}$

$x_{solvent} = \frac{130}{9 + 130} = \frac{130}{139}$

By Raoult's law:

$Δ T_{b} = m \times k_{b}$

$P_{solution} = P_{solvent}^{\circ} x_{solvent}$

$= 760 \times \frac{130}{139} mmHg$

$= 710.8 mmHg \approx 711 mmHg$

Note: $P_{solvent}^{\circ}$ is not given in question, question is solved by taking it as 760 mmHg which is not correct as 760 mmHg is vapor pressure of water at $100 ° C$ but temperature in question is $25 ° C$ .

Q 5 :

2.7 Kg of each of water and acetic acid are mixed. The freezing point of the solution will be $- x$ °C. Consider the acetic acid does not dimerise in water, nor dissociates in water. $x$ = _______ (nearest integer)

[Given: Molar mass of water = 18 g ${mol}^{- 1}$ , acetic acid = 60 g ${mol}^{- 1}$

$K_{f} H_{2} O : 1.86 K k g m o l^{- 1}$

$K_{f} acetic acid : 3.90 K kg {mol}^{- 1}$

$Freezing point: H_{2} O = 273 K, acetic acid = 290 K$ ] [2024]

(31)

$Moles of water = \frac{given mass}{molar mass} = \frac{2700}{18} = 150 mol$

$Moles of acetic acid = \frac{given mass}{molar mass} = \frac{2700}{60} = 45 mol$

As water is in excess, it is solvent.

$Molality (m) = \frac{n_{acetic acid}}{W_{water} (in kg)} = \frac{45}{2.7} m = 16.667 m$

$Δ T_{f} = m k_{f} (solvent) = m k_{f} (water)$

$= 16.667 \times 1.86 ° C = 31 ° C$

$T_{f} (water) - T_{f} (solution) = 31 ° C$

$0 ° C - T_{f} (solution) = 31 ° C$

$T_{f} (solution) = - 31 ° C$

Q 6 :

An artificial cell is made by encapsulating 0.2 M glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a 0.05 M solution of NaCl at 300 K is ______ $\times 10^{- 1}$ bar. (Nearest Integer)

[Given: R = 0.083 L bar ${mol}^{- 1}$ $K^{- 1}$ ]

Assume complete dissociation of NaCl [2024]

(25)

Osmotic pressure developed = Difference of osmotic pressure due to glucose solution difference of osmotic pressure due to NaCl solution.

$= C_{1} R T - i C_{2} R T$

$= 0.2 R T - 2 \times 0.05 R T$

$= 0.1 R T = 0.1 mol \times 0.083 \frac{bar L}{mol K} \times 300 K = 25 \times 10^{- 1} bar$

Q 7 :

Considering acetic acid dissociates in water, its dissociation constant is $6.25 \times 10^{- 5}$ . If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at $- x \times 10^{- 2}$ °C, provided pure water freezes at 0 °C. $x$ = ________ .(Nearest integer)

Given:    ${(K_{f})}_{water}$ = 1.86 K kg ${mol}^{- 1}$ .

density of acetic acid is 1.2 g ${mol}^{- 1}$ .

molar mass of water = 18 g ${mol}^{- 1}$ .

molar mass of acetic acid = 60 g ${mol}^{- 1}$ .

density of water = 1 g ${cm}^{- 3}$

  Acetic acid dissociates as

   ${CH}_{3} COOH ⇌ {CH}_{3} {COO}^{⊝} + H^{\oplus}$ [2024]

(19)

$W_{{CH}_{3} COOH} = density \times volume = 1.2 \frac{g}{mL} \times 5 mL = 6 g$

$W_{H_{2} O} = density \times volume = 1 \frac{g}{mL} \times 1000 mL = 1000 g$

$m = \frac{W_{{CH}_{3} COOH} \times 1000}{M_{{CH}_{3} COOH} \times W_{water} (in g)} = \frac{6 \times 1000}{60 \times 1000 (in g)} m = 0.1 m$

${CH}_{3} COOH ⇌ {CH}_{3} {COO}^{-} + H^{+}$

$K_{a} = \frac{C α^{2}}{1 - α} \approx C α^{2}$

$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{6.25 \times 10^{- 5}}{0.1}} = 0.025$

$i = 1 + (n - 1) α = 1 + (2 - 1) \times 0.025 = 1.025$

$Δ T_{f} = i m k_{f} = 1.025 \times 0.1 \times {1.86}^{\circ} C = {0.19}^{\circ} C$

$T_{f} ° - T_{f} = {0.19}^{\circ} C$

$0 - T_{f} = {0.19}^{\circ} C$

$T_{f} = - {0.19}^{\circ} C = 19 \times {10^{- 2}}^{\circ} C$

Q 8 :

Consider the dissociation of the weak acid HX as given below:

$HX(aq) ⇌ H^{+} (aq) + X^{-} (aq), K a = 1.2 \times 10^{- 5}$

[ $K_{a}$ : dissociation constant]

The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is ______ $\times 10^{- 2}$ bar (nearest integer).

[Given: R = 0.083 L bar ${mol}^{- 1}$ $K^{- 1}$ ] [2024]

(76)

$HX(aq) ⇌ H^{+} (aq) + X^{-} (aq)$

$(t = 0)$ $C$

$(c o n c .)$

$(t = \infty)$ $C (1 - α)$ $C α$ $C α$

$(c o n c .)$

$K_{a} = \frac{C α^{2}}{1 - α} \approx C α^{2}$

$1.2 \times 10^{- 5} = 0.03 α^{2}$

$α = \sqrt{\frac{1.2 \times 10^{- 5}}{0.03}} = 0.02$

$i = 1 + (n - 1) α = 1 + (2 - 1) \times 0.02 = 1.02$

$π = i C R T = 1.02 \times 0.03 \frac{mol}{L} \times 0.083 \frac{bar L}{mol K} \times 300 K = 76.2 \times 10^{- 2} bar$

Q 9 :

0.05M ${CuSO}_{4}$ when treated with 0.01M $K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:

[SPM: Semi Permeable Membrane]

Due to osmosis: [2024]

Green colour formation observed on side X.
Molarity of $K_{2} {Cr}_{2} O_{7}$ solution is lowered.
Green colour formation observed on side Y.
Molarity of ${CuSO}_{4}$ solution is lowered.

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(4)

Solute particles do not move across SPM, so reaction between $K_{2} {Cr}_{2} O_{7}$ and ${CuSO}_{4}$ does not occur and green colour formation does not occur in either chamber. Net movement of solvent is from hypotonic (side X) to hypertonic (side Y) solution. Hence molarity of ${CuSO}_{4}$ solution decreases.

Q 10 :

When $‘ x ’ \times 10^{- 2}$ mL methanol (molar mass = 32 g; density = 0.792 $g / {cm}^{3}$ ) is added to 100 mL water (density = 1 $g / {cm}^{3}$ ), the following diagram is obtained.

x = ___________ (nearest integer)

[Given: Molal freezing point depression constant of water at 273.15 K is 1.86 K kg ${mol}^{- 1}$ ] [2024]

(543)

$Δ T_{f} = m k_{f}$

$(273.15 - 270.65) K = m \times 1.86 {K kg mol}^{- 1}$

$m = \frac{2.5}{1.86} m$

$\frac{W_{methanol} \times 1000}{M_{methanol} \times W_{water} (in g)} = \frac{2.5}{1.86}$

$\frac{(d_{methanol} \times V_{methanol}) \times 1000}{M_{methanol} \times (d_{water} \times V_{water})} = \frac{2.5}{1.86}$

$\frac{(0.792 {g m L}^{- 1} \times x \times 10^{- 2} mL) \times 1000}{32 {g mol}^{- 1} \times (1 {gm L}^{- 1} \times 100 mL)} = \frac{2.5}{1.86} {mol kg}^{- 1}$

$x = 543$

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]

(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:

$M_{A} + 9.874 = 12.346$

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

Q 20 :

In the depression of freezing point experiment

A. Vapour pressure of the solutions is less than that of pure solvent.

B. Vapour pressure of the solutions is more than that of pure solvent.

C. Only solute molecules solidify at the freezing point.

D. Only solvent molecules solidify at the freezing point.

Choose the most appropriate answer from the options given below: [2023]

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

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(4)

In the depression of freezing point experiment

A. Vapour pressure of the solution is less than that of pure solvent

D. Only solvent molecule get solidify.

Q 21 :

Match List I and List II. [2023]

List I

List II

A. Osmosis

I. Solvent molecules pass through semipermeable membrane towards solvent side.

B. Reverse osmosis

II. Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.

C. Electro-osmosis

III. Solvent molecules pass through semipermeable membrane towards solution side.

D. Electrophoresis

IV. Dispersion medium moves in an electric field.

Choose the correct answer from the options given below:

A-I, B-III, C-II, D-IV
A-III, B-I, C-IV, D-II
A-I, B-III, C-IV, D-II
A-III, B-I, C-II, D-IV

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(2)

(i) Electroosmosis: When movement of colloidal particles is prevented by some suitable means (porous diaphragm or semipermeable membranes), it is observed that the D.M. begins to move in an electric field. This phenomenon is termed electroosmosis.

(ii) Solvent molecules pass through semi-permeable membrane towards solvent side is termed as reverse osmosis.

(iii) When an electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied electric potential is called electrophoresis.

(iv) Solvent molecules pass through semipermeable membrane towards the solution side is termed as osmosis.

Q 22 :

Match List I with List II.   [2023]

List I

List II

A. van’t Hoff factor, $i$

I. Cryoscopic constant

B.   $k_{f}$

II.  Isotonic solutions

C. Solutions with same osmotic pressure

III. $\frac{Normal molar mass}{Abnormal molar mass}$

D. Azeotropes

IV. Solutions with same composition of vapour above it

Choose the correct answer from the options given below:

A–III, B–I, C–IV, D–II
A–I, B–III, C–II, D–IV
A–III, B–II, C–I, D–IV
A–III, B–I, C–II, D–IV

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(4)

(A) Van't Hoff factor $= {\frac{Normal molar mass}{Abnormal molar mass}}$

(B) $K_{f} \Rightarrow$ Cryoscopic constant.

(C) Solutions with same osmotic pressure $\Rightarrow$ Isotonic solution.

(D) Azeotropes $\Rightarrow$ Solutions with same composition of vapour above it.

Q 23 :

Evaluate the following statements for their correctness.

(A) The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea.

(B) Azeotropic mixtures boil without change in their composition.

(C) Osmosis always takes place from hypertonic to hypotonic solution.

(D) The density of 32% $H_{2} {SO}_{4}$ solution having molarity 4.09 M is approximately 1.26 g ${mL}^{- 1}$ .

(E) A negatively charged sol is obtained when $KI$ solution is added to silver nitrate solution.

Choose the correct answer from the options given below: [2023]

B and D only
B, D and E only
A and C only
A, B and D only

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(1)

(A) $Δ T_{b}$ is not same, as $' i'$ value is different for two solutions.

(B) Azeotropic mixture: Liquid mixtures which distill over without changes in composition are called constant boiling mixtures or azeotropes or azeotropic mixtures.

(C) Osmosis takes place from hypotonic to hypertonic.

(D) $M = \frac{32 \times 10 \times 1.26}{98} \approx 4.09 M$

(E) When $KI$ is added to ${AgNO}_{3}$ solution then positively charged sol is formed.

Q 24 :

The osmotic pressure of solution of PVC in cyclohexanone at 300 K are plotted on the graph. The molar mass of PVC is ______ g ${mol}^{- 1}$ (Nearest integer).

(Given: R = 0.083 L atm $K^{- 1}$ ${mol}^{- 1}$ ) [2023]

(41500)

Assuming $π$ vs C graph, then slope = $\frac{R T}{M}$

$= \frac{0.083 \times 300}{M} = 6 \times 10^{- 4}$ or $M = 41500 g/mol$

Q 25 :

The number of pairs of the solutions having the same value of the osmotic pressure from the following is _____. (Assume 100% ionization) [2023]

A.   $0.500 M C_{2} H_{5} OH (aq) and 0.25 M KBr(aq)$

B.   $0.100 M K_{4} [Fe {(CN)}_{6}] (aq)$ $and$ $0.100 M {FeSO}_{4} {({NH}_{4})}_{2} {SO}_{4} (aq)$

C.   $0.05 M K_{4} [Fe {(CN)}_{6}] (aq)$ $and$ $0.25 M NaCl (aq)$

D.   $0.15 M NaCl (aq)$ $and$ $0.1 M {BaCl}_{2} (aq)$

E.   $0.02 M KCl . {MgCl}_{2} . 6 H_{2} O (aq)$ $and$ $0.05 M KCl (aq)$

(4)

$π \propto i C$

A, B, D and E

Q 26 :

A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ______ g ${mol}^{- 1}$ . (Nearest Integer)

Given, water boils at 373 K, $K_{b}$ for water = 0.52 K kg ${mol}^{- 1}$ . [2023]

(100)

$Δ T_{b} = k_{b} m$

$(373.52 - 373) = 0.52 \times (\frac{2 \times 1000}{M \times 20})$

$M = \frac{2 \times 1000}{20} = 100 g/mol$

$Molecular mass of solute = 100 g/mol .$

Q 27 :

At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ______ g ${mol}^{- 1}$ .

(Given: R = 0.083 L bar $K^{- 1}$ ${mol}^{- 1}$ ) [2023]

(62250)

$π = C R T$

$\frac{400 \times 10^{- 5}}{(1.01325)} = [\frac{2.5 \times 1000}{M_{solute} \times 250}] (0.083 \times 300)$

$400 \times 10^{- 5} = \frac{10}{M_{solute}} \times 0.083 \times 300$

$M_{solute} = \frac{10 \times 0.083 \times 300}{400} \times 10^{5}$

$= [\frac{10 \times 0.083 \times 3}{4}] \times 10^{5} = 62250 gram/mole$

Q 28 :

25 mL of an aqueous solution of KCl was found to required 20 mL of 1 M ${AgNO}_{3}$ solution when titrated using $K_{2} {CrO}_{4}$ as an indicator. What is the depression in freezing point of KCl solution of the given concentration? (Nearest integer). (Given: $K_{f}$ = 2.0 K kg ${mol}^{- 1}$ )

Assume

(1) 100% ionization and

(2) density of the aqueous solution as 1 g ${mL}^{- 1}$ [2023]

(3)

${25 mL KCl v/s 20 mL of 1M AgNO}_{3}$

$Molarity of KCl = \frac{20 \times 1}{25} = \frac{4}{5} M = \frac{0.8 mol}{L}$

$1 L solution has 0.8 mol KCl$

$Δ T_{f} = i k_{f} m$

$= 2 \times 2 \times 0.8 = 3.2 \approx 3$

Q 29 :

20% of acetic acid is dissociated when its 5 g is added to 500 mL of water. The depression in freezing point of such water is ______ $\times 10^{- 3}$ °C. Atomic mass of C, H and O are 12, 1 and 16 a.m.u. respectively. (Given: Molal depression constant and density of water are 1.86 K kg ${mol}^{- 1}$ and 1 g ${cm}^{- 3}$ respectively) [2023]

(372)

$α = 0.2$

$i = 1 + (n - 1) α = 1.2$

$Δ T_{f} = i k_{f} m = 1.2 \times 1.86 \times \frac{5 \times 1000}{60 \times 500}$

$= 0.372 = 372 \times 10^{- 3} ° C$

Q 30 :

Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is/are __________. [2023]

A. 1 M aq. NaCl and 2 M aq. urea

B. 1 M aq. ${CaCl}_{2}$ and 1.5 M aq. KCl

C. 1.5 M aq. ${AlCl}_{3}$ and 2 M aq. ${Na}_{2} {SO}_{4}$

D. 2.5 M aq. KCl and 1 M aq. ${Al}_{2} {({SO}_{4})}_{3}$

(4)

For isotonic solution $i_{1} c_{1} = i_{2} c_{2}$

Solution	$i_{1} c_{1}$	$i_{2} c_{2}$
A	2	2
B	3	3
C	6	6
D	5	5

List I	List II
A. Osmosis	I. Solvent molecules pass through semipermeable membrane towards solvent side.
B. Reverse osmosis	II. Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.
C. Electro-osmosis	III. Solvent molecules pass through semipermeable membrane towards solution side.
D. Electrophoresis	IV. Dispersion medium moves in an electric field.

Topic Question Set

Q 1 :

What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Q 2 :

The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

Q 3 :

Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

(Molar mass in g ${mol}^{- 1}$ for ethylene glycol 62, $K_{f}$ of water = 1.86 K kg ${mol}^{- 1}$ ) [2024]

Q 9 :

0.05M ${CuSO}_{4}$ when treated with 0.01M $K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:

[SPM: Semi Permeable Membrane]

Due to osmosis: [2024]

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]

Q 24 :

The osmotic pressure of solution of PVC in cyclohexanone at 300 K are plotted on the graph. The molar mass of PVC is ______ g ${mol}^{- 1}$ (Nearest integer).

(Given: R = 0.083 L atm $K^{- 1}$ ${mol}^{- 1}$ ) [2023]

Q 26 :

A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ______ g ${mol}^{- 1}$ . (Nearest Integer)

Given, water boils at 373 K, $K_{b}$ for water = 0.52 K kg ${mol}^{- 1}$ . [2023]

Q 27 :

At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ______ g ${mol}^{- 1}$ .

(Given: R = 0.083 L bar $K^{- 1}$ ${mol}^{- 1}$ ) [2023]

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List I	List II
A. van’t Hoff factor, $i$	I. Cryoscopic constant
B. $k_{f}$	II. Isotonic solutions
C. Solutions with same osmotic pressure	III. $\frac{Normal molar mass}{Abnormal molar mass}$
D. Azeotropes	IV. Solutions with same composition of vapour above it

Topic Question Set

Q 1 : What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Q 2 : The osmotic pressure of a dilute solution is 7×105 Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ ×104 Nm-2. [2024]

Q 3 : Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg. (Molar mass in g mol-1 for ethylene glycol 62, Kf of water = 1.86 K kg mol-1) [2024]

Q 8 : Consider the dissociation of the weak acid HX as given below: HX(aq)⇌H+(aq)+X-(aq), Ka=1.2×10-5 [Ka: dissociation constant] The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is ______ ×10-2 bar (nearest integer). [Given: R = 0.083 L bar mol-1 K-1] [2024]

Q 9 : 0.05M CuSO4 when treated with 0.01M K2Cr2O7 gives green colour solution of Cu2Cr2O7. The two solutions are separated as shown below: [SPM: Semi Permeable Membrane] Due to osmosis: [2024]

Q 10 : When ‘x’×10-2 mL methanol (molar mass = 32 g; density = 0.792 g/cm3) is added to 100 mL water (density = 1 g/cm3), the following diagram is obtained. x = ___________ (nearest integer) [Given: Molal freezing point depression constant of water at 273.15 K is 1.86 K kg mol-1] [2024]

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]

Q 24 : The osmotic pressure of solution of PVC in cyclohexanone at 300 K are plotted on the graph. The molar mass of PVC is ______ g mol-1 (Nearest integer). (Given: R = 0.083 L atm K-1 mol-1) [2023]

Q 26 : A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ______ g mol-1. (Nearest Integer) Given, water boils at 373 K, Kb for water = 0.52 K kg mol-1. [2023]

Q 27 : At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ______ g mol-1. (Given: R = 0.083 L bar K-1 mol-1) [2023]

Q 30 : Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is/are __________. [2023] A. 1 M aq. NaCl and 2 M aq. urea B. 1 M aq. CaCl2 and 1.5 M aq. KCl C. 1.5 M aq. AlCl3 and 2 M aq. Na2SO4 D. 2.5 M aq. KCl and 1 M aq. Al2(SO4)3

Want Us to Email you About Special Offers & Updates?

Q 1 :

What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Q 2 :

The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

Q 3 :

Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

(Molar mass in g ${mol}^{- 1}$ for ethylene glycol 62, $K_{f}$ of water = 1.86 K kg ${mol}^{- 1}$ ) [2024]

Q 9 :

0.05M ${CuSO}_{4}$ when treated with 0.01M $K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:

[SPM: Semi Permeable Membrane]

Due to osmosis: [2024]

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]

Q 24 :

The osmotic pressure of solution of PVC in cyclohexanone at 300 K are plotted on the graph. The molar mass of PVC is ______ g ${mol}^{- 1}$ (Nearest integer).

(Given: R = 0.083 L atm $K^{- 1}$ ${mol}^{- 1}$ ) [2023]

Q 26 :

A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ______ g ${mol}^{- 1}$ . (Nearest Integer)

Given, water boils at 373 K, $K_{b}$ for water = 0.52 K kg ${mol}^{- 1}$ . [2023]

Q 27 :

At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ______ g ${mol}^{- 1}$ .

(Given: R = 0.083 L bar $K^{- 1}$ ${mol}^{- 1}$ ) [2023]