Solutions — Page 6 | NEET Chemistry

Q51

The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 torr partial pressure is :

A 0.0075

B 0.015

C 0.02

D 0.005

Correct Answer

Option B

Solution

Partial pressure = Mole fraction × Solubility

{{{p_1}} \over {{p_2}}} = {{{s_1}} \over {{s_2}}}

$\Rightarrow$

{{500} \over {750}} = {{0.01} \over {{s_2}}}

$\Rightarrow$ s2 = 0.015 g L-1

Q52

In the depression of freezing point experiment A. Vapour pressure of the solution is less than that of pure solvent B. Vapour pressure of the solution 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 :

A A and C only

B A only

C A and D only

D B and C only

Correct Answer

Option C

Solution

In the depression of freezing point experiment only solvent molecules solidify and vapour pressure of solution decreases as some of the surface area is occupied by solute molecules, so less number of molecules will go in vapour form.

Q53

1.24 g of AX2 (molar mass 124 g mol−1) is dissolved in 1 kg of water to form a solution with boiling point of 100.015°C, while 25.4 g of AY2 (molar mass 250 g mol−1) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.Kb(H2O) = 0.52 kg mol−1Which of the following is correct?

A AX2 and AY2 (both) are completely unionised.

B AX2 and AY2 (both) are fully ionised.

C AX2 is completely unionised while AY2 is fully ionised.

D AX2 is fully ionised while AY2 is completely unionised.

Correct Answer

Option D

Solution

Mass of $A x_2=1.24 \mathrm{~g}$ (solute) Molarmass of $A X_2=124 \mathrm{~g} \mathrm{~mol}^{-1}$ Mass of water $=1 \mathrm{~kg}$ (solvent.)

Boiling point of water $=100^{\circ} \mathrm{C}$ Boiling point of water after adding solute $A X_2=100.0156^{\circ} \mathrm{C}$ Mass of $A Y_2=25.4 \mathrm{~g}$ (solute) Molarmass of $A Y_2=250 \mathrm{~g~mol}^{-1}$ Mass of water $=2 \mathrm{~kg}$ (Solvent) Boiling point of water $\mp 100^{\circ} \mathrm{C}$ Boiling point of water after adding solute $A y_2=100.0260^{\circ} \mathrm{C}$ $\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1}=0.520^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}$ The ionisation of $A x_2$ and $A Y_2$ can be determined by calculating Van't Hoff factor.

The phenomenon given in the question is elevation in boiling point.

The boiling point of solvent increases when another compound (solute) is added to it.

Relation: $\Delta T_b=K_b \cdot i \cdot m$

\Delta T_b=T_b-T_b^0

$\Delta T_b \rightarrow$ Boiling point elevation $T_b \rightarrow$ Boiling point of solution (solvent + solute) $T_b^0 \rightarrow$ Boiling point of solvent $K_b \rightarrow$ Molal elevation constant $i \rightarrow$ Van't Hoff factor

m \rightarrow \text{ Molality }=\frac{\text{ Number of moles }}{\mathrm{kg} \text{ of solvent }}, \text{ Moles }=\frac{\text{ Mass }}{\text{ Molarmass }}

For $A X_2$ and $A Y_2$ (solute), Van't Hoff factor represents how many particles a solute dissociates into when dissolved in a solvent (water).

For non-electrolytes, $i=1$

\begin{aligned} &\begin{aligned} & A X_2: \\ & \text{ Moles }=\frac{\text{ mass }}{\text{ molarmass }}=\frac{1.249}{124 \mathrm{~g~mol^{-1}}}=0.01 \mathrm{~mol} \\ & \text{ Molality }=\frac{\text{ Moles }}{k_g \text{ of solvent }}=\frac{0.01 \mathrm{~mol}}{1 \mathrm{~kg}}=0.01 \mathrm{~mol} \mathrm{~kg}^{-1} \\ & \Delta T_b=K_b \times i \times m \\ & i=\frac{\Delta T_b}{k_b \times m} \\ &=\frac{T_b-T_b^0}{K_b \times m} \end{aligned}\\ &\text{ Substitute values as., } \end{aligned}

\begin{aligned} i & =\frac{\left(100.0156^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}\right)}{0.52^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.01 \mathrm{~mol} \mathrm{~kg}^{-1}} \\ & =\frac{0.0156^\circ \mathrm{C}}{0.52 \times 0.01{ }^{\circ} \mathrm{C}}=3 \end{aligned}

$i=3$ means, there are 3 particles in solution after $A X_2$ dissolved in water.

A x_2 \rightarrow 1 A^{+}+2 X^{-} \quad(1+2=3)

\begin{aligned} &A x_2 \text{ is completely ionised. }\\ &A Y_2: \end{aligned}

\begin{aligned} &\begin{aligned} \text{ Moles } & =\frac{\text{ Mass }}{\text{ Molarmass }}=\frac{25.4 \mathrm{~g}}{250 \mathrm{g~mol}}=0.1016 \mathrm{~mol} \\ \text{ Molality } & =\frac{\text{ Moles }}{\mathrm{kgof}^{-1} \text{ solvent }}=\frac{0.1016 \mathrm{~mol}}{2 \mathrm{~kg}}=0.050 .8 \mathrm{~mol} \mathrm{~kg}^{-1} \\ \Delta T_b & =k_b \times \mathrm{l}^2 \times \mathrm{m} \\ i & =\frac{\Delta T_b}{k_b \times m} \\ & =\frac{T_b-T_b^0}{k_b \times m} \end{aligned}\\ &\text{ Substitute values as, } \end{aligned}

i = {{(100.0260^\circ C - 100^\circ C)} \over {0.52^\circ C\,kg\,mo{l^{ - 1}} \times 0.0508\,mol\,k{g^{ - 1}}}}

$=\dfrac{0.0260 ^\circ \mathrm{C}}{0.52 \times 0.0508 ^\circ \mathrm{C}}$ $= 0.98$ $\approx 1$ $i=1$ means $A Y_2$ is completely unionised. $A y_2$ not give ionised particles when dissolved in water.

So, $A X_2$ is completely ionised and $A Y_2$ is completely unionised.

Answer: Option 4) $A x_2$ is fully ionised, $A Y_2$ is completely unionised.

Q54

The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following processes can explain this?

A Osmosis

B Reverse osmosis

C Diffusion

D Dialysis

Correct Answer

Option A

Solution

Raw mango shrink in salt solution due to net transfer of water molecules from mango to salt solution due to phenomenon of osmosis.

Q55

Equimolar solutions in the same solvent have

A Same boiling point but different freezing point

B Same freezing point but different boiling point

C Same boiling and same freezing points

D Different boiling and different freezing points

Correct Answer

Option C

Solution

Equimolar solutions of normal solutes in the same solvent will have the same boiling point and same freezing point.

Q56

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6oC will be : [Kf for water = 1.86 K kg mol−1 , and molar mass of ethylene glycol = 62 g mol−1 )

A 204.30 g

B 400.00 g

C 304.60 g

D 804.32 g

Correct Answer

Option D

Solution

Given

{K_r} = 1.86\,K\,kg\,mo{l^{ - 1}}

\Delta {T_f} = 0 - \left( { - 6} \right) = {6^ \circ }C

As we know that

\Delta {T_f} = {K_f} \times \,\,molality

= {{{K_f} \times 1000 \times mass\,\,of\,\,solute} \over {molar\,\,mass\,\,of\,\,solute\,\, \times \,\,mass\,\,of\,\,solvent\,\,in\,\,kg}}

Substituting given values in formula

6 = {{1.86 \times 1000 \times w} \over {62 \times 4}};

\,\,\,\,\,w = 0.8\,kg = 800\,gm

Q57

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

Q58

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%.

Q59

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

Q60

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.