Solutions — Page 5 | JEE Chemistry

Q41

The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L–1) in solution is :

A 4 × 10–4

B 6 × 10–2

C 4 × 10–2

D 16 × 10–4

Correct Answer

Option B

Solution

Given that, osmotic pressure of XY(

{\pi _{XY}}

) in water is four times that of a solution of 0.01 M BaCl2(

{\pi _{BaC{l_2}}}

). $\therefore$

{\pi _{XY}}

= 4 $\times$

{\pi _{BaC{l_2}}}

$\Rightarrow$ 2 [XY] = 4 $\times$ 3 $\times$ 0.01 $\Rightarrow$ [XY] = 0.06 = 6 × 10–2 mol L–1

Q42

Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx =

{3 \over 4}

My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :

A

{4 \over 3}

B

{3 \over 4}

C

{1 \over 2}

D

{1 \over 4}

Correct Answer

Option B

Solution

Relative lowering of vapour pressure,

{{\Delta P} \over P}

=

{{{n_2}} \over {{n_1}}}

n2 = Number of moles of solute n1 = Number of moles of solvent.

Given that, Here is 5 molal solution, means 5 moles of solute are dissolved in 1 kg or 1000 g of solvent.

\therefore\,\,\,

Number of moles of solute = 5 Number of moles of solvent X =

{{1000} \over {{M_X}}}

Number of moles of solvent Y =

{{1000} \over {My}}

\therefore\,\,\,

{\left( {{{\Delta \,P} \over P}} \right)_x}

=

{5 \over {{{1000} \over {{M_x}}}}}

=

{{5{M_x}} \over {1000}}

{\left( {{{\Delta \,P} \over P}} \right)_y}

=

{5 \over {{{1000} \over {{M_y}}}}}

=

{{5\,{M_y}} \over {1000}}

Accoding to the question.

{{5{M_x}} \over {1000}}

= m $\times$

{{5\,{M_y}} \over {1000}}

$\Rightarrow$

\,\,\,

Mx = m $\times$ My $\Rightarrow$

\,\,\,

{3 \over 4}

My = m $\times$ My [as given, Mx =

{3 \over 4}

My] $\Rightarrow$

\,\,\,

m =

{3 \over 4}

Q43

Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is

A Kb = Kf

B Kb = 0.5 Kf

C Kb = 1.5 Kf

D Kb = 2 Kf

Correct Answer

Option D

Solution

{{\Delta {T_b}} \over {\Delta {T_f}}} = {{i.m \times {k_b}} \over {i \times m \times {k_f}}}

{2 \over 2} = {{1 \times 1 \times {k_b}} \over {1 \times 2 \times {k_f}}}

{k_b} = 2{k_f}

Q44

The freezing point of benzene decreases by 0.450C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (Kf for benzene = 5.12 K kg mol–1)

A 80.4 %

B 74.6 %

C 94.6 %

D 64.6 %

Correct Answer

Option C

Solution

\Delta

Tf = i $\times$ Kf $\times$ m $\Rightarrow$ 0.45 = i $\times$ 5.12 $\times$

{{0.2 \times 1000} \over {60 \times 20}}

$\Rightarrow$ i = 0.527 2CH3COOH ⇌ (CH3COOH)2 1 - $\alpha$

{\alpha \over 2}

i = 1 - $\alpha$ +

{\alpha \over 2}

$\Rightarrow$ $\alpha$ = 0.946 $\therefore$ % dissociation is 94.6%.

Q45

Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vaapor pressures of pure A and pure B are 7

\times

103 Pa and 12

\times

103 Pa, respectively . The composition of the vapor in equilibriumwith a solution containing 40 mole percent of A at this temperature is :

A xA = 0.76; xB = 0.24

B xA = 0.28; xB = 0.72

C xA = 0.4; xB = 0.6

D xA = 0.37; xB = 0.63

Correct Answer

Option B

Solution

{y_A} = {{{P_A}} \over {{P_{Total}}}} = {{P_A^0{x_A}} \over {P_A^0{X_A} \times P_B^0X{}_B}}

= {{7 \times {{10}^3} \times 0.4} \over {7 \times {{10}^3} \times 0.4 + 12 \times {{10}^3} \times 0.6}}

= {{2.8} \over {10}} = 0.28

{y_B} = 0.72

Q46

Given below are two statements : 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 :

A Both Statement I and Statement II are false

B Statement I is true but Statement II is false

C Both Statement I and Statement II are true

D Statement I is false but Statement II is true

Correct Answer

Option C

Solution

Statement I : Correct NaCl addition to ice causes preventing the melting of ice.

On adding NaCl to ice, freezing point lowers.

This creates a colder mixture, preventing the ice cream from melting.

Melting point of ice is 0 $^\circ$ C.

When only ice is used to make ice cream, at 0 $^\circ$ C ice starts melting by absorbing the energy from its environment in the form of heat.

Addition of salt to ice while making the cream lowers the freezing point of the ice, allowing it to reach a colder temperature and thus the ice cream mixture freezes properly, So, the salt causes ice to melt at a lower temperature than pure ice.

Statement II : Correct Decrease in freezing point while addition of NaCl to ice at 0 $^\circ$ C is due to the colligative property depression in freezing point.

So, both the statements are correct.

Both statement I and statement II are true.

Q47

The freezing point of a diluted milk sample is found to be –0.2oC, while it should have been –0.5oC for pure milk. How much water has been added to pure milk to make the diluted sample?

A 1 cup of water to 2 cups of pure milk

B 2 cups of water to 3 cups of pure milk

C 3 cups of water to 2 cups of pure milk

D 1 cup of water to 3 cups of pure milk

Correct Answer

Option C

Solution

We know,

\Delta

Tf = i $\times$ kf $\times$ m $\therefore$

\Delta

TfDil. Milk = (1) $\times$ kf $\times$ mdil = 0.2 ...(1)

\Delta

TfPure Milk = (1) $\times$ kf $\times$ mpure = 0.5 ...(2) From (1) and (2), we get

{{{m_{pure}}} \over {{m_{dil}}}} = {{0.5} \over {0.2}}

=

{5 \over 2}

....(

3) Assume moles of fat in both the milk samples = n Weight of milk = w1 and weight of water = w0 Molality of pure milk, mpure =

{n \over {{w_1}}} \times 1000

....(4) Molality of diluted milk, mdil. =

{n \over {{w_1} + {w_0}}} \times 1000

....(5) From (4) and (5), we get

{{{m_{pure}}} \over {{m_{dil}}}} = {{{w_1} + {w_0}} \over {{w_1}}}

Using equation (3), we get

{5 \over 2} = {{{w_1} + {w_0}} \over {{w_1}}}

$\Rightarrow$ 5w1 = 2w1 + 2w0 $\Rightarrow$ 3w1 = 2w0 $\Rightarrow$

{{{w_1}} \over {{w_0}}} = {2 \over 3}

Which means 3 cups of water has been added to 2 cups of pure milk.

Q48

Match List I with List II .tg .tg List I List II A. van't Hoff factor, i I. Cryoscopic constant B.

\mathrm{k_f}

II. Isotonic solutions C. Solutions with same osmotic pressure III.

\mathrm{\dfrac{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 A-III, B-I, C-IV, D-II

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

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

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

Correct Answer

Option B

Solution

.tg .tg A. van't Hoff factor III.

\mathrm{{{Normal\,molar\,mass} \over {Abnormal\,molar\,mass}}}

B.

\mathrm{k_f}

I.

Cryoscopic constant C.

Solutions with same osmotic pressure II.

Isotonic solutions D.

Azeotropes IV.

Solutions with same composition of vapour above it

Q49

Given below are two statements: Statement (I) : Molal depression constant

\mathrm{K}_f

is given by

\dfrac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\mathrm{fus}}}

, where symbols have their usual meaning. Statement (II) :

\mathrm{K}_f

for benzene is less than the

\mathrm{K}_f

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

A Statement I is correct but Statement II is incorrect

B Both Statement I and Statement II are incorrect

C Statement I is incorrect but Statement II is correct

D Both Statement I and Statement II are correct

Correct Answer

Option A

Solution

Statement-I Molar depression constant $\mathrm{k}_f=\dfrac{\mathrm{M}_1 \mathrm{RT}_f^2}{\Delta \mathrm{H}_{\text{fus }}}$

\begin{aligned} & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_{\mathrm{f}}}{\left[\frac{\Delta \mathrm{H}_{\mathrm{fus}}}{\mathrm{~T}_{\mathrm{f}}}\right]} \\ & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\text{fus }}} \end{aligned}

Hence statement-I is correct but $\mathrm{k}_{\mathrm{f}}$ for benzene $=5.12 \dfrac{{ }^{\circ} \mathrm{C}}{\text{ molal }}$ $\mathrm{k}_{\mathrm{f}}$ for water $=1.86 \dfrac{{ }^{\circ} \mathrm{C}}{\text{ molal }}$ Hence statement- II is incorrect

Q50

\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{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 isGiven:

K_f

(H2O) = 1.8 K kg mol−1, molality ≡ molarity

A

1.90 \times 10^{-3}

B

1.38 \times 10^{-3}

C

1.1 \times 10^{-2}

D

1.89 \times 10^{-1}

Correct Answer

Option B

Solution

Freezing Point Depression: $\Delta T_f = i \times K_f \times m$ Given: $\Delta T_f = 0.2 \, \text{°C}, \quad K_f = 1.8 \, \text{K kg mol}^{-1}, \quad m = 0.1 \, \text{m}$ Substituting the given values: $0.2 = i \times 1.8 \times 0.1$ Solving for $i$ : $i = \dfrac{0.2}{1.8 \times 0.1} = \dfrac{20}{18} = \dfrac{10}{9}$ Degree of Dissociation ( $\alpha$ ): For the reaction $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$ : $i = 1 + \alpha$ Given $i = \dfrac{10}{9}$ : $\dfrac{10}{9} = 1 + \alpha$ $\alpha = \dfrac{1}{9}$ Dissociation Constant ( $K_{eq}$ ): $\mathrm{K}_{eq} = \dfrac{[H^+][A^-]}{[HA]}$ At equilibrium: $[H^+] = [A^-] = \alpha \times C = \dfrac{1}{9} \times 0.1$ $[HA] = 0.1 \times (1 - \alpha) = 0.1 \times \left(1 - \dfrac{1}{9}\right)$ Substituting these into $K_{eq}$ : $\mathrm{K}_{eq} = \dfrac{(0.1 \times \dfrac{1}{9})^2}{0.1 \times \left(1 - \dfrac{1}{9}\right)}$ Simplifying: $\mathrm{K}_{eq} = \dfrac{0.1 \times \left(\dfrac{1}{81}\right)}{0.1 \times \dfrac{8}{9}} = \dfrac{1}{720}$ Therefore: $\mathrm{K}_{eq} = 1.38 \times 10^{-3}$