Solutions MCQs for NEET — Chemistry Questions with Answers

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

Key Idea: Vapour pressure depends upon the surface area of the solution. Larger the surface area, higher is the vapour pressure.

Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles, resulting in decreased surface area. However, addition of solvent, i.e., dilution, increases the surface area of the liquid surface, thus results in increased vapour pressure.

Hence, addition of water to the aqueous solution of (1 molal) KI, results in increased vapour pressure.

Depression in freezing point

$∆$T_f = k_f x m

where, m = molality = w_B x 1000/M_B.W_A = 68.5 x 1000/342 x 1000 = 68.5/342

 $∆$T_f = 1.86 x 68.5/342 = 0.372$°$C

$∆$T_f = T$°$-T_s = 0-0.372$°$C

Key Idea ${\mathrm{\lambda }}_{\mathrm{eq}} = \mathrm{k} \mathrm{x} \mathrm{v} = \left(\mathrm{kx}1000\right)\mathrm{normality}$

On dilution, the number of current carrying particles per cm³ decreases but the volume of solution increases. Consequently, the ionic mobility increases, which in turn increases the equivalent conductance of strong electrolyte.

Degree of dissociation, α=^^c/^∞ where, ^^c and ^∞ are equivalent conductances at a given concentration and at infinite dilution respectively.
    ⇒ α=8.0/400=2x10^-2

From Ostawald’s dilution law (for weak monobasic acid)

K_a = Cα/(1- α) = Cα² (∴ 1>>>α) = 1/32 (2x10^-2)² = 1.25x10^-5