Question

An aqueous solution $10\%$ by mass of organic compound $\left(C_{12}{H}_{16}{N}_2{O}_2\right)$ freezes at $-{1.8}^{\circ }C$. Then select the correct statements (s):
[Given: $K_f$ (water) = $1.8Kkgmo{l}^{-1}$]

• A. Van't hoff factor is 1.98
• B. The vapour pressure of solution will be less than vapour pressure of pure water
• C. This compound associates in water
• D. This compound dissociates in water

Answer 1

Answer: (A), (B), (D)

The correct options are
A Van't hoff factor is 1.98
B The vapour pressure of solution will be less than vapour pressure of pure water
D This compound dissociates in water

(A) Given mass of organic compound
= 10 g
So, mass of solvent = 100 - 10 = 90 g
$K_f\left(water\right)=18Kkgmo{l}^{-1}$ and
$T_f=-{1.8}^{\circ }C\Delta T_f=i\times K_f\times m$
$[\text{molarity}=\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{[mass of solvent (kg)]}}]1.8=i\times 1.8\times \frac{10}{220}\times \frac{1000}{90}$
$i=1.98$
($i>1$. So, compound dissociates)
(B) Vapour pressure of solution having non-volatile solute has less vapour pressure than pure solvent.
(C) As ($i>1$. So, compound dissociates)