Solutions — Page 2 | NEET Chemistry

Q11

The mixture which shows positive deviation from Raoult's law is :

A Benzene + Toluene

B Acetone + Chloroform

C Chloroethane + Bromoethane

D Ethanol + Acetone

Correct Answer

Option D

Solution

Pure ethanol molecules are hydrogen bonded.

On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds between them.

This weakness the intermolecular attractive interactions and the solution shows positive deviation from Raoult's law.

Q12

For an ideal solution, the correct option is :

A

\Delta

mix H = 0 at constant T and P

B

\Delta

mix G = 0 at constant T and P

C

\Delta

mix S = 0 at constant T and P

D

\Delta

mix V

\ne

0 at constant T and P

Correct Answer

Option A

Solution

For an ideal solution at constant T and P,

\Delta

mix H = 0,

\Delta

mix S > 0,

\Delta

mix G < 0 and

\Delta

mix V = 0.

Q13

The mixture that forms maximum boiling azeotrope is :

A Ethanol + Water

B Acetone + Carbon disulphide

C Heptane + Octane

D Water + Nitric acid

Correct Answer

Option D

Solution

Maximum boiling azeotrope is shown by solution which shows negative deviation from Raoult's law.

Except water + Nitric acid, all other mixtures show negative deviation.

Q14

If molality of the dilute solution is doubled, the value of molal depression constant ( K f ) will be

A halved

B tripled

C unchanged

D doubled.

Correct Answer

Option C

Solution

The value of molal depression constant, K f is constant for a particular solvent, thus, it will be unchanged when molality of the dilute solution is doubled.

Q15

The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is

A 0

B 1

C 2

D 3

Correct Answer

Option D

Solution

Ba(OH) 2 is a strong electrolyte and undergoes cent percent dissociation in a dilute aqueous solution.

Ba(OH) 2(aq) $\to$ Ba 2+ (aq) + 2OH – (aq) Thus, van’t Hoff factor i = 3.

Q16

Which one of the following is incorrect for ideal solution?

A

\Delta {H_{mix}} = 0

B

\Delta {U_{mix}} = 0

C

\Delta P = {P_{obs}} - P

calculated by Raoult's law

D

\Delta {G_{mix}} = 0

Correct Answer

Option D

Solution

For ideal solution, we have

\Delta

H mix = 0,

\Delta

V mix = 0 Now U mix = ∆H mix – P

\Delta

V mix $\therefore$

\Delta

U mix = 0 Also, for an ideal solution, p A = x A p A o , p B = x B p B o $\therefore$

\Delta

p = p observed – p calculated = 0

\Delta

G mix =

\Delta

H mix – T

\Delta

S mix For an ideal solution,

\Delta

S mix $\ne$ 0 $\therefore$

\Delta

G mix $\ne$ 0

Q17

At 100 o C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K b = 0.52, the boiling point of this solution will be

A 102 o C

B 103 o C

C 101 o C

D 100 o C

Correct Answer

Option C

Solution

Given that w s = 6.5 g, w A = 100 g p s = 732 mm of Hg k b = 0.52, T o b = 100 o C p o = 760 mm of Hg

{{{p^o} - {p_s}} \over {{p^o}}} = {{{n_2}} \over {{n_1}}}

$\Rightarrow$

{{760 - 732} \over {760}} = {{{n_2}} \over {{{100} \over {18}}}}

$\Rightarrow$ n 2 = 0.2046 mol

\Delta

T b = K b × m T b - T o b =

{k_b} \times {{{n_2} \times 1000} \over {{w_{A\left( g \right)}}}}

$\Rightarrow$ T b - 100 o C =

{{0.52 \times 0.2046 \times 1000} \over {100}}

= 1.06 $\Rightarrow$ T b = 101.06 o C

Q18

Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25 o C. (Given, vapour pressure data at 25 o C, benzene = 12.8 kPa, toluene = 3.85 kPa)

A The vapour eill contain equal amounts of benzene and toluene.

B Not enough information is given to make a prediction.

C The vapour will contain a higher percentage of benzene.

D The vapour will contain a higher percentage of toluene

Correct Answer

Option C

Solution

p Benzene = x Benzene . p o Benzene p Toluene = x Toluene . p o Toluene For an ideal 1 : 1 molar mixture of benzene and toluene x Benzene =

{1 \over 2}

and x Toluene =

{1 \over 2}

p Benzene =

{1 \over 2}

p o Benzene =

{1 \over 2} \times 12.8

= 6.4 kPa p Toluene =

{1 \over 2}

p o Toluene =

{1 \over 2} \times 3.85

= 1.925 kPa Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene.

Q19

Which one of the following electrolytes has the same value of van't Hoff factor (i) as that of Al 2 (SO 4 ) 3 (if all are 100% ionised)?

A Al(NO 3 ) 3

B K 4 [Fe(CN) 6 ]

C K 2 SO 4

D K 3 [Fe(CN) 6 ]

Correct Answer

Option B

Solution

Ai 2 (SO 4 ) 3 ⇌ 2Al +3 + 3SO 4 2– van't Hoff factor, i = 5 K 2 SO 4 ⇌ 2K + + SO 4 2– van't Hoff factor, i = 3 K 3 [Fe(CN) 6 ] ⇌ 3K + + [Fe(CN) 6 ] 2- van't Hoff factor, i = 4 Al(NO 3 ) 3 ⇌ Al 3+ + 3NO 3 – van't Hoff factor, i = 4 K 4 [Fe(CN) 6 ] ⇌ 4K + + [Fe(CN) 6 ] 4– van't Hoff factor, i = 5

Q20

The boiling point of 0.2 mol kg

-

1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?

A Molecular mass of X is less than the molecular mass of Y .

B Y is undergoing dissociation in water while X undergoes no change.

C X is undergoing dissociation in water.

D Molecular mass of X is greater than the molecular mass of Y .

Correct Answer

Option C

Solution

\Delta

T b = iK b m Given, (

\Delta

T b ) x > (

\Delta

T b ) y $\therefore$ i x K b m > i y K b m $\Rightarrow$ i x > i y (K b is same for same solvent) So, x is undergoing dissociation in water.