Solutions — Page 8 | NEET Chemistry

Q71

Which one of the following statements regarding Henry's law is not correct ?

A Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids

B Different gases have different KH (Henry's law constant) values at the same temperature.

C The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.

D The value of KH increases with increase of temperature and KH is function of the nature of the gas

Correct Answer

Option A

Solution

From Henry's law we know, Pgas = KH xg Where, Pgas = Pressure of undissolved gas xg = Mole fraction of gas dissolved into the liquid.

KH = Henry's Carnot.

When pressure is constant then, xg $\propto$

{1 \over {K{}_H}}

So. when KH is high then xg or solubility of gas is lower in the liquid. So, option (A) is wrong.

Q72

Among the following mixtures, dipole-dipole as the major interaction, is present in

A benzene and ethanol

B acetonitrile and acetone

C KCl and water

D benzene and carbon tetrachloride

Correct Answer

Option B

Solution

Acetonitrile

\mathop {\left( {C{H_3}} \right.}\limits^{\delta + } \,

\,\, - \,\,\,C \equiv \,\,

\mathop {\left. N \right)}\limits^{\delta - }

and acetone dipole-dipole interaction exist between them. Between

KCl

and water ion-dipole interaction is found and in Benzene - ethanol, Benzene is non polar but ethanol is polar so it create dipole - induced dipole interaction Benzene $-$ Carbon tetra chloride, both are non polar so one create instantaneous dipole and other gets induced by it.

Q73

Molecules of benzoic acid (C6H5COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1)

A 1.5 g

B 1.8 g

C 1.0 g

D 2.4 g

Correct Answer

Option D

Solution

We know,

\Delta

Tf = i Kf $\times$

{w \over M} \times {{1000} \over {{w_s}}}

i = 1 + $\alpha$

\left( {{1 \over n} - 1} \right)

Here Benzoic acid dimerise, so value of n = 2 $\therefore$ i = 1 + 0.8

\left( {{1 \over 2} - 1} \right)

= 1 $-$ 0.4 = 0.6 $\therefore$ 2 = 0.6 $\times$ 5 $\times$

{w \over {122}} \times {{1000} \over {30}}

$\Rightarrow$ w = 2.44 g

Q74

Boiling point of a

2 \%

aqueous solution of a non-volatile solute A is equal to the boiling point of

8 \%

aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is

A

\mathrm{M}_{\mathrm{A}}=4 \mathrm{M}_{\mathrm{B}}

B

\mathrm{M}_{\mathrm{B}}=4 \mathrm{M}_{\mathrm{A}}

C

\mathrm{M}_{\mathrm{A}}=8 \mathrm{M}_{\mathrm{B}}

D

\mathrm{M}_{\mathrm{B}}=8 \mathrm{M}_{\mathrm{A}}

Correct Answer

Option B

Solution

For $\mathbf{A}: 100 \,\mathrm{gm}$ solution $\rightarrow 2 \,\mathrm{gm}$ solute $\mathrm{A}$ $\therefore$ Molality $=\dfrac{2 / \mathrm{M}_{\mathrm{A}}}{0.098}$ For B : $100 \,\mathrm{gm}$ solution $\rightarrow 8 \,\mathrm{gm}$ solute $\mathrm{B}$

\begin{aligned} &\therefore \text{ Molality }=\frac{8 / \mathrm{M}_{\mathrm{B}}}{0.092} \\\\ &\because\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{A}}=\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{B}} \end{aligned}

$\therefore$ Molality of $\mathrm{A}=$ Molality of $\mathrm{B}$

\therefore \frac{2}{0.098 \mathrm{M}_{\mathrm{A}}}=\frac{8}{0.092 \mathrm{M}_{\mathrm{B}}}

\frac{2}{98} \times \frac{92}{8}=\frac{M_A}{M_B}

\frac{1}{4.261}=\frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}}

\therefore \mathrm{M}_{\mathrm{B}}=4.261 \times \mathrm{M}_{\mathrm{A}}

Q75

A solution containing 62 g ethylene glycol in 250 g water is cooled to

-

10oC. If Kf for water is 1.86 K kg mol

-

1 , the amount of water (in g) separated as ice is :

A 48

B 32

C 64

D 16

Correct Answer

Option C

Solution

Here water is solvent and ethylene glycol is solute. We know, Depression of freezing point,

\Delta

Tf = Kf . m

\Delta {T_f} = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{250} \over {1000}}} \right)}}

= 7.44 We know, freezing point of pure water (here solvent) is 0oC. Which is represented by

T_f^0

. We also know,

\Delta {T_f} = T_f^0 - {T_f}

where

T_f^0

= Freezing point of pure solvent and

{T_f}

= Freezing point of solution $\therefore$ 7.44 = 0 -

{T_f}

$\Rightarrow$

{T_f}

= -7.44oC So, not a single drop of water solution will become ice until temperature reaches -7.44oC.

When temperature decrease more than -7.44oC then some part of the solution starts becoming ice staring from surface of the solution.

Now let Wl gm of solution still stays in liquid phase when temperature reaches -10oC.

$\therefore$

10 = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{{W_l}} \over {1000}}} \right)}}

Wl = 186 gm So, The amount of water (in g) separated as ice is

\Delta

W = (250 $-$ 186) = 64 gm

Q76

The solution from the following with highest depression in freezing point/lowest freezing point is

A

180 \mathrm{~g}

of acetic acid dissolved in benzene

B

180 \mathrm{~g}

of acetic acid dissolved in water

C

180 \mathrm{~g}

of benzoic acid dissolved in benzene

D

180 \mathrm{~g}

of glucose dissolved in water

Correct Answer

Option B

Solution

\Delta \mathrm{T}_{\mathrm{f}}

is maximum when

\mathrm{i} \times \mathrm{m}

is maximum. 1)

\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha

Hence

\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha 2)

\mathrm{m}_2=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=7.68^{\circ} \mathrm{C}

3)

\mathrm{m}_3=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=3.8^{\circ} \mathrm{C}

4)

\mathrm{m}_4=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}_{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}{ }^{\prime}=1.86^{\circ} \mathrm{C}

As per NCERT,

\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}

\mathrm{k}_{\mathrm{f}}{ }^{\prime}(\text { Benzene })=5.12 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$$

Q77

Arrange the following solutions in order of their increasing boiling points. (i)

10^{-4} \mathrm{M} \mathrm{NaCl}

(ii)

10^{-4} \mathrm{M}

Urea (iii)

10^{-3} \mathrm{M} \mathrm{NaCl}

(iv)

10^{-2} \mathrm{M} \mathrm{NaCl}

A

(

i

)<(

ii

)<(

iii

)<(

iv

)

B (ii)

<(

i

)<(

iii

)<(

iv

)

C (iv)

<(

iii

)<(

i

)<(

ii

)

D (ii)

<

(i)

\equiv

(iii)

<

(iv)

Correct Answer

Option B

Solution

Step 1: Identify the van’t Hoff factor ( $i$ ) for each solute NaCl dissociates (ideally) into two ions: $\mathrm{NaCl} \;\rightarrow\; \mathrm{Na^+} + \mathrm{Cl^-},$ so $i \approx 2.$ Urea ( $\mathrm{CH_4N_2O}$ ) is a non‐electrolyte (does not dissociate), so $i = 1.$ Step 2: Effective molar concentration of particles The total particle concentration for each solution is approximately $(i \times \text{molarity})$ . (i) $10^{-4}\,M$ NaCl $\text{Effective concentration} \;=\; 2 \times 10^{-4} = 2 \times 10^{-4}.$ (ii) $10^{-4}\,M$ Urea $\text{Effective concentration} \;=\; 1 \times 10^{-4} = 1 \times 10^{-4}.$ (iii) $10^{-3}\,M$ NaCl $\text{Effective concentration} \;=\; 2 \times 10^{-3} = 2 \times 10^{-3}.$ (iv) $10^{-2}\,M$ NaCl $\text{Effective concentration} \;=\; 2 \times 10^{-2} = 2 \times 10^{-2}.$ Step 3: Compare to rank the boiling points A larger total particle concentration (and hence larger colligative effect) corresponds to a higher boiling point.

Arrange from lowest to highest: Lowest: $10^{-4}\,M$ Urea $\bigl[1 \times 10^{-4}\bigr]$ Next: $10^{-4}\,M$ NaCl $\bigl[2 \times 10^{-4}\bigr]$ Next: $10^{-3}\,M$ NaCl $\bigl[2 \times 10^{-3}\bigr]$ Highest: $10^{-2}\,M$ NaCl $\bigl[2 \times 10^{-2}\bigr]$ Hence, in the format $(\text{ii}) Final Answer$ \boxed{\text{(ii) }

Q78

Which one of the following 0.06 M aqueous solutions has lowest freezing point?

A Al2(SO4)3

B C6H12O6

C KI

D K2SO4

Correct Answer

Option A

Solution

To determine which solution has the lowest freezing point, you should consider the van 't Hoff factor $i$ , which represents the number of particles the solute dissociates into when dissolved.

The freezing point depression is given by:

\Delta T_f = i \cdot K_f \cdot m

where: $\Delta T_f$ is the change in freezing point, $K_f$ is the freezing point depression constant, $m$ is the molality of the solution, $i$ is the van 't Hoff factor, which depends on the dissociation of the solute.

For each solute, the van 't Hoff factor $i$ can be determined as follows: Option A: $\text{Al}_2(\text{SO}_4)_3$ Dissociates into 2 Al $^{3+}$ ions and 3 SO $_4^{2-}$ ions.

$i = 2 + 3 = 5$ Option B: $\text{C}_6\text{H}_{12}\text{O}_6$ (glucose) Does not dissociate in solution (non-electrolyte).

$i = 1$ Option C: $\text{KI}$ Dissociates into K $^+$ and I $^-$ ions.

$i = 1 + 1 = 2$ Option D: $\text{K}_2\text{SO}_4$ Dissociates into 2 K $^+$ ions and 1 SO $_4^{2-}$ ion.

$i = 2 + 1 = 3$ Comparing the van 't Hoff factors, the solution of $\text{Al}_2(\text{SO}_4)_3$ will have the highest factor ( $i = 5$ ), leading to the greatest freezing point depression.

Thus, the $\text{Al}_2(\text{SO}_4)_3$ solution will have the lowest freezing point since the extent of freezing point depression is directly proportional to $i$ .

Therefore, the solution of $\text{Al}_2(\text{SO}_4)_3$ has the lowest freezing point.

Q79

18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :

A 76.0

B 752.4

C 759.0

D 7.6

Correct Answer

Option B

Solution

According to Raoult's Law

{{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{W_B} \times {M_A}} \over {{M_B} \times {W_A}}}\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Here

{P^ \circ } =

Vapour pressure of pure solvent,

{P_s} =

Vapour pressure of solution

{W_B} =

Mass of solute,

{W_A} =

Mass of solvent

{M_B} =

Molar mass of solute,

{M_A} =

Molar Mass of solvent Vapour pressure of pure water at

{100^ \circ }C

(by assumption

=760

torr) By substituting values in equation

(i)

we get,

{{760 - {P_s}} \over {{P_s}}} = {{18 \times 18} \over {180 \times 178.2}}\,\,\,\,\,\,...\left( {ii} \right)

On solving

(ii)

we get On solving

(ii)

we get

{P_s} = 752.4\,\,torr

Q80

Freezing point of an aqueous solution is (-0.186)oC. Elevation of boiling point of the same solution is Kb = 0.512 oC, Kf = 1.86 oC, find the increase in boiling point.

A 0.186 oC

B 0.0512 oC

C 0.092 oC

D 0.2732 oC

Correct Answer

Option B

Solution

\Delta {T_b} = {K_b}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;

\Delta {T_f} = {K_f}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;

{{\Delta {T_b}} \over {\Delta {T_f}}} = {{{K_b}} \over {{K_f}}} = {{\Delta {T_b}} \over { - 0.186}}

= {{0.512} \over {1.86}}

= {0.0512^ \circ }C.