Question

0.534 g of solute is dissolved in 15 g of water then freezing point temperature changes from $0^{o}C$ to $-{1.57}^{o}C$. Molal depression constant of water, $K_f=1.85Kkgmo{l}^{-1}$. Find out:
(i) Molal concentration
(ii) Molecular mass of solute

• A. (i)$0.848$ (ii) $54.98$ amu
• B. (i)$0.748$ (ii) $41.98$ amu
• C. (i)$0.848$ (ii) $41.98$ amu
• D. none

Answer 1

The correct option is C (i)$0.848$ (ii) $41.98$ amu

we have,

$\Delta T_f=K_f\ast m$

1.57 = 1.85 * m

molal concentration = 0.848 m

m = $\frac{givenmass}{molarmass\ast massofsolute}$

0.848 = $\frac{0.534\ast 1000}{x\ast 15}$

molar mass(x) = 41.98 amu