MHT CET 2019 Chemistry Question Paper with Answer and Solution

(D) The relationship between the van’t Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is given by:
$\alpha = \frac{i - 1}{n' - 1}$
Where,$n'$ is the number of ions formed after dissociation.
For the dissociation reaction: $A \rightleftharpoons n'B$
Initially: $1$ mole,$0$
After dissociation: $(1 - \alpha)$ mole,$n'\alpha$
Total number of moles present in the solution:
$= (1 - \alpha) + n'\alpha = 1 + (n' - 1)\alpha$
By definition,the van’t Hoff factor $(i)$ is the ratio of observed colligative property to calculated colligative property,which equals the ratio of total moles after dissociation to initial moles:
$i = \frac{1 + (n' - 1)\alpha}{1} = 1 + (n' - 1)\alpha$
Rearranging for $\alpha$:
$i - 1 = (n' - 1)\alpha$
$\alpha = \frac{i - 1}{n' - 1}$

(A) The dissociation of $Ba(NO_3)_2$ is given by: $Ba(NO_3)_2 \rightleftharpoons Ba^{2+} + 2NO_3^{-}$.
Here,the number of ions produced per formula unit is $n = 3$.
The formula for the degree of dissociation $(\alpha)$ in terms of the Van't Hoff factor $(i)$ is: $\alpha = \frac{i-1}{n-1}$.
Substituting the given values: $\alpha = \frac{2.74-1}{3-1} = \frac{1.74}{2} = 0.87$.

(B) Given: Molality $(m) = 0.2 \ m$,Freezing point of solution $(T_f) = -0.680 \ ^{\circ}C$,Freezing point of pure water $(T_f^{\circ}) = 0 \ ^{\circ}C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Depression in freezing point $(\Delta T_f) = T_f^{\circ} - T_f = 0 - (-0.680) = 0.680 \ K$.
The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Substituting the values: $0.680 = i \times 1.86 \times 0.2$.
$i = \frac{0.680}{1.86 \times 0.2} = \frac{0.680}{0.372} \approx 1.8279$.
Rounding to two decimal places,the Van't Hoff factor $(i) \approx 1.83$.

(D) The precipitation power of an electrolyte increases with the charge of the flocculating ion. This is explained by the $Hardy-Schulze$ rule,which states that the greater the valence (charge) of the flocculating ion added,the greater is its power to cause precipitation.