Question

Solution containing 0.73 g camphor (Molar mass $152$ g mol${}^{-1})$ in 36.8 g of acetone (boiling point ${56.3}^{\circ }C$ boil at ${56.55}^{\circ }C$. A solution of 0.564 g unknown compound in same weight of acetone boils at ${56.46}^{\circ }C.$ Calculate the molar mass of the unknown compound.

Answer 1

Calculation for $K_b$ of camphor

$W_A=36.8g$, $W_B=0.73g$

$M_B=152$

$\Delta T_b=56.55-56.30={0.25}^{o}C$

Now, $\Delta T_b=K_b\times \frac{W_B\times 1000}{W_A\times M_B}$

$\therefore K_b=\frac{\Delta T_b{W}_A{M}_B}{W_B\times 1000}=\frac{0.25\times 36.8\times 152}{0.73\times 1000}$

$={1.9156}^{o}Ckg/mo{l}^{-1}$

Calculation for molecular mass of unknown solute

$W_A=36.8g,W_B=0.564g$

$\Delta T_b=56.46-56.3={0.16}^{o}C$

Now, $M_B=\frac{K_b{W}_{B}1000}{W_A\times \Delta T_b}=\frac{1.9156\times 0.564\times 1000}{36.8\times 0.16}$

$\Rightarrow M_B=183.4g$