Solutions Questions — NEET Chemistry

Q1

Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to :

A Increase in escaping tendency of molecules of each component.

B Formation of hydrogen bonding between acetone and chloroform

C Stronger intermolecular forces between chloroform molecules than those between chloroform and acetone molecules.

D Repulsive forces.

Correct Answer

Option B

Solution

Acetone and chloroform show negative deviation from Raoult's law due to stronger H-bonding between acetone and chloroform mixture.

Hence, escaping tendency decreases ∴ Vapour pressure decreases ∴ Boiling point increases.

Q2

5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure

X

and

Y

are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?

A The solution is ideal.

B The solution has volume greater than the sum of individual volumes.

C The solution shows positive deviation.

D The solution shows negative deviation.

Correct Answer

Option D

Solution

Given: 5 moles of liquid X 10 moles of liquid Y The total moles in the solution are 15 (5 moles X + 10 moles Y).

The mole fraction of X ( $X_x$ ) is: $\dfrac{5}{15} = \dfrac{1}{3}$ The mole fraction of Y ( $X_y$ ) is: $\dfrac{10}{15} = \dfrac{2}{3}$ The vapor pressures of pure liquids are: $P_x^{\circ} = 63$ torr for liquid X $P_y^{\circ} = 78$ torr for liquid Y Using Raoult's Law for an ideal solution, the total vapor pressure ( $P_{\text{total}}$ ) is calculated as: $P_{\text{total}} = X_x \cdot P_x^{\circ} + X_y \cdot P_y^{\circ}$ Substitute the values: $P_{\text{total}} = \left(\dfrac{1}{3} \times 63\right) + \left(\dfrac{2}{3} \times 78\right)$ Calculate each term: $= 21 + 52$ $= 73 \text{ torr}$ The observed vapor pressure of the solution is 70 torr, which is lower than the calculated ideal pressure of 73 torr.

This indicates a negative deviation from Raoult's Law because the observed vapor pressure is less than expected for an ideal solution.

Therefore, the solution exhibits stronger intermolecular attractions than those in an ideal solution, leading to a lower vapor pressure.

Q3

Mass of glucose

(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6)

required to be dissolved to prepare one litre of its solution which is isotonic with

15 \mathrm{~g} \mathrm{~L}^{-1}

solution of urea

(\mathrm{NH}_3 \mathrm{CONH}_2)

is (Given: Molar mass in

\mathrm{g} \mathrm{mol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{~N}: 14

)

A 55 g

B 15 g

C 30 g

D 45 g

Correct Answer

Option D

Solution

For isotonic solutions [osmotic pressure must be equal]

\begin{aligned} & \pi_1=\pi_2 \\ & C_1 R T=C_2 R T \\ & C_1=C_2 \\ & \frac{\mathrm{m}}{180 \times 1}=\frac{15}{60 \times 1} \quad \text{ ( } \mathrm{m} \text{ is the mass of glucose) } \\ & \mathrm{m}=\frac{180}{4}=45 \mathrm{~g} \end{aligned}

Q4

The Henry's law constant

(\mathrm{K}_{\mathrm{H}})

values of three gases

(\mathrm{A}, \mathrm{B}, \mathrm{C})

in water are

145, 2 \times 10^{-5}

and

35 \mathrm{~kbar}

respectively. The solubility of these gases in water follow the order:

A B > A > C

B B > C > A

C A > C > B

D A > B > C

Correct Answer

Option B

Solution

According to Henry's law, the solubility of a gas in a liquid under constant temperature is directly proportional to the partial pressure of the gas above the liquid but is inversely proportional to the Henry's law constant

(\mathrm{K}_\mathrm{H})

for the gas. Henry's law can be expressed as:

\mathrm{c} = \frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}

where

\mathrm{c}

is the concentration (or solubility) of the gas in the liquid,

\mathrm{P}

is the partial pressure of the gas, and

\mathrm{K}_{\mathrm{H}}

is Henry's law constant. From the equation, it is clear that the solubility of the gas is inversely related to

\mathrm{K}_{\mathrm{H}}

; if

\mathrm{K}_{\mathrm{H}}

increases, the solubility decreases, and vice versa. In the given problem, you have the

\mathrm{K}_{\mathrm{H}}

values of the gases A, B, and C as follows: A: 145 kbar B:

2 \times 10^{-5} \mathrm{~kbar}

C: 35 kbar Comparing the

\mathrm{K}_{\mathrm{H}}

values: Gas B has the lowest

\mathrm{K}_{\mathrm{H}}

and hence the highest solubility. Gas A, with the highest

\mathrm{K}_{\mathrm{H}}

among the three, will have the lowest solubility. Gas C has a

\mathrm{K}_{\mathrm{H}}

value less than A but greater than B, so its solubility will be lower than B but higher than A.

Therefore, the order of solubility of the gases in water from the highest to the lowest is: B > C > A Thus, the correct option is: Option B: B > C > A

Q5

The plot of osmotic pressure (П) vs concentration

(\mathrm{mol} \mathrm{~L}^{-1})

for a solution gives a straight line with slope

25.73 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1}

. The temperature at which the osmotic pressure measurement is done is (Use

\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

)

A

37^{\circ} \mathrm{C}

B

310^{\circ} \mathrm{C}

C

25.73^{\circ} \mathrm{C}

D

12.05^{\circ} \mathrm{C}

Correct Answer

Option A

Solution

The relationship between the osmotic pressure

(\Pi)

of a solution and its concentration

(c)

can be derived from the van't Hoff equation for dilute solutions, which is given by:

\Pi = cRT

where:

\Pi

is the osmotic pressure,

c

is the concentration of the solution in moles per liter,

R

is the ideal gas constant in appropriate units, and

T

is the temperature in Kelvin. According to the problem, the slope of the

\Pi

vs

c

plot is given as

25.73 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1}

which corresponds to the product

RT

from the van't Hoff equation. We are provided with the value of the gas constant

R = 0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

. To find the temperature

T

, we use the following equation derived from the slope of the line:

RT = 25.73 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1}

To isolate

T

, we rearrange the equation:

T = \frac{25.73 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1}}{0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}} \approx 310 \mathrm{~K}

To convert this temperature from Kelvin to Celsius, we use the conversion formula:

T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15

T_{\text{Celsius}} = 310 \mathrm{~K} - 273.15 \approx 36.85^{\circ} \mathrm{C}

This value is closest to

37^{\circ} \mathrm{C}

, which corresponds to Option A.

Therefore, the temperature at which the osmotic pressure measurement is done is approximately

37^{\circ} \mathrm{C}

.

Q6

Which amongst the following aqueous solution of electrolytes will have minimum elevation in boiling point? Choose the correct option :-

A 0.05 M NaCl

B 0.1 M KCl

C 0.1 M MgSO

_4

D 1 M NaCl

Correct Answer

Option A

Solution

<p>

\mathrm{i\times M\downarrow \Rightarrow \Delta T_b \downarrow}

</p> <p> <table class=tg style=undefined;table-layout: fixed; width: 299px> <colgroup> <col style=width: 152px> <col style=width: 147px> </colgroup> <thead> <tr> <th class=tg-amwm>Electrolyte</th> <th class=tg-amwm>

\mathrm{i\times M}

</th> </tr> </thead> <tbody> <tr> <td class=tg-0lax>

\mathrm{NaCl}

</td> <td class=tg-0lax>

2\times0.05=0.1

</td> </tr> <tr> <td class=tg-0lax>

\mathrm{KCl}

</td> <td class=tg-0lax>

2\times0.1=0.2

</td> </tr> <tr> <td class=tg-0lax>

\mathrm{MgSO_4}

</td> <td class=tg-0lax>

2\times0.1=0.2

</td> </tr> <tr> <td class=tg-0lax>

\mathrm{NaCl}

</td> <td class=tg-0lax>

2\times1=2

</td> </tr> </tbody> </table></p>

Q7

<p>K<sub>H</sub> value for some gases at the same temperature 'T' are given :</p> <p> <table class=tg style=undefined;table-layout: fixed; width: 279px> <colgroup> <col style=width: 144px> <col style=width: 135px> </colgroup> <thead> <tr> <th class=tg-baqh>Gas</th> <th class=tg-baqh>K

_H

/k bar</th> </tr> </thead> <tbody> <tr> <td class=tg-baqh>Ar</td> <td class=tg-baqh>40.3</td> </tr> <tr> <td class=tg-baqh>CO

_2

</td> <td class=tg-baqh>1.67</td> </tr> <tr> <td class=tg-baqh>HCHO</td> <td class=tg-baqh>1.83

\times

10

^{ - 5}

</td> </tr> <tr> <td class=tg-baqh>CH

_4

</td> <td class=tg-baqh>0.413</td> </tr> </tbody> </table></p> <p>where K<sub>H</sub> is Henry's Law constant in water. The order of their solubility in water is :</p>

A HCHO 4 2 < Ar

B Ar 2 4 < HCHO

C Ar 4 2 < HCHO

D HCHO 2 4 < Ar

Correct Answer

Option B

Solution

According to Henry's Law, p = K H x Where 'p' is partial pressure of gas in vapour phase.

K H is Henry's Law constant. 'x' is mole fraction of gas in liquid.

Higher the value of K H at a given pressure, lower is the solubility of the gas in the liquid.

$\therefore$ Solubility : Ar 2 4

Q8

The following solutions were prepared by dissolving 10 g of glucose (C 6 H 12 O 6 ) in 250 ml of water (P 1 ), 10 g of urea (CH 4 N 2 O) in 250 ml of water (P 2 ) and 10 g of sucrose (C 12 H 22 O 11 ) in 250 ml of water (P 3 ). The right option for the decreasing order of osmotic pressure of these solutions is :

A P 3 > P 1 > P 2

B P 2 > P 1 > P 3

C P 1 > P 2 > P 3

D P 2 > P 3 > P 1

Correct Answer

Option B

Solution

$\bullet$ Osmotic pressure ( $\pi$ ) = iCRT where C is molar concentration of the solution $\bullet$ With increase in molar concentration of solution osmotic pressure increases.

$\bullet$ Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.

$\bullet$ Order of molar mass of solute decreases as Sucrose > Glucose > Urea $\bullet$ So, correct order of osmotic pressure of solution is P 3 > P 1 > P 2

Q9

The correct option for the value of vapour pressure of a solution at 45

^\circ

C with benzene to octane in molar ratio 3 : 2 is : [At 45

^\circ

C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

A 350 mm of Hg

B 160 mm of Hg

C 168 mm of Hg

D 336 mm of Hg

Correct Answer

Option D

Solution

Given :

{n_{{C_6}{H_6}}}:{n_{{C_8}{H_{18}}}} = 3:2

So,

{\chi _{{C_6}{H_6}}} = {3 \over 5},{\chi _{{C_8}{H_{18}}}} = {2 \over 5}

Total vapour pressure of solution,

{p_s} = p_{{C_6}{H_6}}^o{\chi _{{C_6}{H_6}}} + p_{{C_8}{H_{18}}}^o{\chi _{{C_8}{H_{18}}}}

= 280 \times {3 \over 5} + 420 \times {2 \over 5}

= 168 + 168

= 336

mm of Hg

Q10

The freezing point depression constant (K r ) of benzene is 5. 12 K kg mol -1 . The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is rounded off upto two decimal places :

A 0.80 K

B 0.40 K

C 0.60 K

D 0.20 K

Correct Answer

Option B

Solution

\Delta {T_f} = {k_f}m

= 5.12 $\times$ 0.078 = 0.399 K = 0.4 K