Solutions — Page 10 | NEET Chemistry

Q91

Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is: (xM = Mole fraction of 'M' in solution ; xN = Mole fraction of 'N' in solution ; yM = Mole fraction of 'M' in vapour phase ; yN = Mole fraction of 'N' in vapour phase)

A

{{{x_M}} \over {{x_N}}} < {{{y_N}} \over {{y_N}}}

B (xM – yM) < (xN – yN)

C

{{{x_M}} \over {{x_N}}} = {{{y_N}} \over {{y_N}}}

D

{{{x_M}} \over {{x_N}}} > {{{y_M}} \over {{y_N}}}

Correct Answer

Option D

Solution

P_M^0

= 450 mmHg and

P_N^0

= 700 mmHg $\therefore$

P_M^0

<

P_N^0

Also we know,

P_M

=

P_M^0

X_M

=

Y_M

P_T

$\Rightarrow$

P_M^0

=

{{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)

P_N

=

P_N^0

X_N

=

Y_N

P_T

$\Rightarrow$

P_N^0

=

{{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)

As,

P_M^0

<

P_N^0

$\Rightarrow$

{{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)

<

{{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)

$\Rightarrow$

{{{Y_M}} \over {{X_M}}}

<

{{{Y_N}} \over {{X_N}}}

$\Rightarrow$

{{{Y_M}} \over {{Y_N}}}

<

{{{X_M}} \over {{X_N}}}

Q92

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be

A 360

B 350

C 300

D 700

Correct Answer

Option B

Solution

P_A^ \circ = ?,\,\,

Given

P_B^ \circ = 200mm,\,\,{x_A} = 0.6

{x_B} = 1 - 0.6 = 0.4,\,\,P = 290

P = {P_A} + {P_B} = P_A^ \circ {x_A} + P_B^ \circ {x_B}

\Rightarrow 290 = P_A^ \circ \times 0.6 + 200 \times 0.4

$\therefore$

\,\,\,\,\,\,\,\,

P_A^ \circ = 350mm

Q93

An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :

A 0.33

B 0.50

C 0.67

D 0.80

Correct Answer

Option B

Solution

Let us assume that degree of dissociation is $\alpha$ .

\begin{array}{lll}{M{X_2}} & { \to {M^{2 + }} + } & {2X + } \\ {(1 - \alpha )} & \alpha & {2\alpha } \end{array}

Thus, after dissociation total number of moles formed (n) = 3. Now, we know degree of dissociation is

\alpha = {{i - 1} \over {n - 1}} = {{2 - 1} \over {3 - 1}} = 0.50

Q94

The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:

A 64

B 128

C 488

D 32

Correct Answer

Option A

Solution

Using relation,

{{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{w_2}{M_1}} \over {{w_1}{M_2}}}

where

{w_1},

{M_1} =

mass in

g

and mol. mass of solvent

{w_2},{M_2} =

mass in

g

and mol. mass of solute Let

{M_2} = x

{P^ \circ } = 185\,\,torr;\,\,{P_s} = 183\,torr

{{185 - 183} \over {183}} = {{1.2 \times 58} \over {100x}}

(Mol. mass of acetone

=58

)

x=64

$\therefore$

\,\,\,\,\,

Molar mass of substance

=64

Q95

Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes?

A

\text{CH}_3\text{OH} + \text{CHCl}_3

B

\text{C}_6\text{H}_5\text{OH} + \text{C}_6\text{H}_5\text{NH}_2

C

\text{H}_2\text{O} + \text{CH}_3\text{COC}_2\text{H}_5

D

\text{CS}_2 + \text{CH}_3\text{COCH}_3

Correct Answer

Option B

Solution

A binary mixture of $\text{C}_6\text{H}_5\text{OH}$ and $\text{C}_6\text{H}_5\text{NH}_2$ exhibits negative deviation from Raoult's law.

This means that the vapor pressure of the solution is lower than the vapor pressures of the pure components, $\text{C}_6\text{H}_5\text{OH}$ and $\text{C}_6\text{H}_5\text{NH}_2$ .

Consequently, the boiling point of the solution is higher than the boiling points of the pure substances.

Therefore, this mixture forms a maximum boiling azeotrope.

Q96

We have three aqueous solutions of

\mathrm{NaCl}

labelled as '

\mathrm{A}

', '

\mathrm{B}

' and '

\mathrm{C}

' with concentration

0.1 \mathrm{M}

,

0.01 \mathrm{M}

and

0.001 \mathrm{M}

, respectively. The value of van 't Hoff factor(i) for these solutions will be in the order :

A

\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{C}}<\mathrm{i}_{\mathrm{B}}

B

\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{B}}<\mathrm{i}_{\mathrm{C}}

C

\mathrm{i}_{\mathrm{A}}>\mathrm{i}_{\mathrm{B}}>\mathrm{i}_{\mathrm{C}}

D

\mathrm{i}_{\mathrm{A}}=\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}}

Correct Answer

Option B

Solution

th, .header { background-color: #0000FF; color: white; } Salt Values of i (for different conc. of a Salt) NaCl 0.1 M 0.01 M 0.001 M 1.87 1.94 1.94 The van 't Hoff factor (i) is used to describe the number of particles a solute formula unit produces in a solution.

For an electrolyte like $\mathrm{NaCl}$ , which dissociates completely in very dilute solutions, the theoretical value of $i$ is approximately 2, since $\mathrm{NaCl}$ dissociates into $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ions.In real scenarios, as the concentration of the solution decreases (making the solution more dilute), the interaction between the ions decreases, allowing more complete dissociation.

Therefore, for practical purposes, the van 't Hoff factor $i$ approaches its theoretical maximum value as concentration decreases.

Thus, for $\mathrm{NaCl}$ solutions of concentrations $0.1 \mathrm{M}$ , $0.01 \mathrm{M}$ , and $0.001 \mathrm{M}$ , the fact that $\mathrm{NaCl}$ dissociates more completely in more dilute solutions implies that $i$ increases with decreasing concentration.Hence, the order of $i$ based on the concentration would be $\mathrm{i}_{\mathrm{A}} 0.01 \mathrm{M} > 0.001 \mathrm{M}$ , respectively.

Thus, option B correctly describes the order of the van 't Hoff factors for these solutions.

Q97

A solution is made by mixing one mole of volatile liquid

A

with 3 moles of volatile liquid

B

. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg . The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

A

1400 \mathrm{~mm} \mathrm{~Hg}, \mathrm{A}

B

1400 \mathrm{~mm} \mathrm{~Hg}, B

C

600 \mathrm{~mm} \mathrm{~Hg}, \mathrm{A}

D

600 \mathrm{~mm} \mathrm{~Hg}, \mathrm{B}

Correct Answer

Option C

Solution

Given: 1 mole of volatile liquid A 3 moles of volatile liquid B Vapor pressure of pure A, $P_A^o = 200$ mm Hg Vapor pressure of the solution, $P_{S} = 500$ mm Hg We apply Raoult's law, which states: $P_{S} = P_A^o \cdot X_A + P_B^o \cdot X_B$ Where: $X_A$ is the mole fraction of A $X_B$ is the mole fraction of B $P_B^o$ is the vapor pressure of pure liquid B Calculate the mole fractions: $X_A = \dfrac{1}{1+3} = \dfrac{1}{4}$ $X_B = \dfrac{3}{1+3} = \dfrac{3}{4}$ Plug these into the equation: $500 = 200 \times \dfrac{1}{4} + P_B^o \times \dfrac{3}{4}$ Simplifying: $500 = 50 + \dfrac{3}{4} P_B^o$ Subtract 50 from both sides: $450 = \dfrac{3}{4} P_B^o$ Multiply both sides by $\dfrac{4}{3}$ to solve for $P_B^o$ : $P_B^o = 600 \, \text{mm Hg}$ Since $P_A^o In conclusion: The vapor pressure of pure B,$ P_B^o $, is 600 mm Hg.

The least volatile component is A.

Q98

For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would be expected to have the same values? (Assume ideal behaviour)

A Heat of vaporization

B Gaseous densities at the same temperature and pressure

C Boiling points

D Vapour pressure at the same temperature

Correct Answer

Option B

Solution

Gaseous densities of ethanol and dimethyl ether would be same at same temperature and pressure.

The heat of vaporisation ,

V.P.

and

b.pts

will differ due to

H

-bonding in ethanol.