Chemistry MCQs for NEET — Practice Questions with Answers

(d) Key Idea FOr this problem, the following expression can be used.

                      $∆{\mathrm{G}}_{\mathrm{mix}}=∆{\mathrm{H}}_{\mathrm{mix}}-\mathrm{T}∆{\mathrm{S}}_{\mathrm{mix}}$

For an ideal gas

                      $∆{\mathrm{H}}_{\mathrm{mix}}= 0; ∆{\mathrm{U}}_{\mathrm{mix}} =0; ∆{\mathrm{S}}_{\mathrm{mix}}\ne 0$

Putting all these values in the expression,

            $$\begin{gathered} ∆{\mathrm{G}}_{\mathrm{mix}}=∆{\mathrm{H}}_{\mathrm{mix}}-\mathrm{T}∆{\mathrm{S}}_{\mathrm{mix}} \\ ∆{\mathrm{G}}_{\mathrm{mix}}= 0-\mathrm{T}∆{\mathrm{S}}_{\mathrm{mix}} \\ \therefore ∆{\mathrm{G}}_{\mathrm{mix}} \ne 0 \end{gathered}$$

Thus, option (d) is incorrect.

(d) From Raoult's law of  paratial pressure,

(p_A-p_s)/p_s = n_B / n_A

= 760-732 / 732 = W_B x M_A / M_B x W_A

=28/732 = 6.05x18 /M_A x 100

= M_A = 30.6

$∆$Tb = 0.52x6.5x1000/30.6x100 = 1.10

Boiling point = 100 + 1.10

=101.1°C=101°C

(d) Since, component having higher vapour pressure will have higher percentage in vapour phase. Benzene has vapour pressure 12.8 kPa which is greater than toluene 3.85 kPa.

Therefore, the vapour will contain a higher percentage of benzene.

For an ideal solution:-
(i)There will be no change in volume on mixing the two components i.e. ΔV_mixing = 0.
(ii)There will be no change in volume on ΔH_mixing = 0. So, ΔS_mix≠0 for an ideal solution.

higher is the dissociation ,higher will be the colligative properties.

When solute undergoes dissociation than vant Hoff factor i>ΔTb=iKbm

Al₂(SO₄)₃⇌2AL³⁺ + 3SO₄^2-

Value of van't Hoff's factor (i)=5

(a) K₂SO₄⇌ 2K⁺ + SO₄^2- (i=3)

(b) K₃[Fe(CN)₆]⇌ 3K⁺ + [Fe(CN)₆]^3- (i=4)

(c) Al(NO₃)₃⇌ Al³⁺ + 3NO^3- (i=4)

(d) K₄[Fe(CN)₆]⇌ 4K⁺ + [Fe(CN)₆]^3- (i=5)

Therefore, K₄[Fe(CN)₆] has same value of i that of Al₂(SO₄)₃ i.e. i=5

$∆$T_f (freezing point depression) is a colligative property and depends upon the van't Hoff factor (i), i.e., number of ions given by the electrolyte in aqueous solution.

$∆$T_f =i x K_f xm

where, ${\mathrm{k}}_{\mathrm{f}}$ = molal freezing point depression constant

m = molality of the solution

K_f and m are constant, $∆{\mathrm{T}}_{\mathrm{f}} \mathrm{\alpha } \mathrm{i}$

(a) KCl(aq) $\leftrightharpoons$ K⁺(aq) + Cl^-(aq),

(Total ions =2 thus, i = 2)

(b) C₆H₁₂O₆ $\rightleftharpoons$no ions[i=0]

(c) Al₂(SO₄)₃(aq) $\leftrightharpoons$2Al³⁺ + 3SO^2-₄ [Total ions = 5, thus, i = 5 ]

(d) K₂SO₄(aq) $\rightleftharpoons$2K⁺ + SO^-

[Total ions = 3, thus, i = 3 ]

Hence, Al₂(SO₄)₃ will exhibit largest freezing point depression due to the highest value of.

Total pressure,
p_T = p'_A+p'_B ...(i)
We know that, p'_A = p_Ax_A
                      p'_B = p_Bx_B
Substituting the values of p'_A and p'_B in Eq. (i)
p_T = p_Ax_A + p_Bx_B

[x_A+x_B=1->x_A=1-x_B or x_B=1-x_A]

=p_Ax_A + p_B(1-x_A) = p_Ax_A + p_B - p_Bx_A

∴ p_T = p_B+x_A(p_A-p_B)

(b) For dissociation, i>1

   For association, i<1