Question

$20 \,g$ of an non-volatile solute is added to $500 \,g$ of solvent, freezing point of pure solvent $=5.48^{\circ} C$ and that of solution is $4.47^{\circ} C , K _{ f }=1.93 \,k \,kg \,mol ^{-1}$ molecular mass of solute is:

• A. 77.2
• B. 76.4
• C. 73.2
• D. 70.6

Answer 1

Answer: (B)

$\Delta T _{ f }=5.48-4.47=1.01, \Delta T _{ f }= K _{ f } \times m$

$1.01=1.93 \times \frac{20 \times 1000}{ m \times 50} \Rightarrow m =76.4$