Question

Boiling point of pure chloroform is ${61}^{\circ }C.$ .
On addition of $5.0g$ of a non-volatile solute to $20g$ pure chloroform , the solution boils at ${64.63}^{\circ }C$
$\text{Ebullioscopic constant}{K}_b=3.63Kkgmo{l}^{-1}$
Calculate the molecular weight of the solute

• A. $320gmo{l}^{-1}$
• B. $100gmo{l}^{-1}$
• C. $400gmo{l}^{-1}$
• D. $250gmo{l}^{-1}$

Answer 1

The correct option is D $250gmo{l}^{-1}$

$\text{Mass of solute}{w}_2=5.0g$
$\text{Mass of solvent}{w}_1=20g$
$K_b=3.63Kkgmo{l}^{-1}$
$\Delta T_b=64.63-61={3.63}^{\circ }C$
Using,
$\Delta T_b=\frac{(k_b\times 1000\times w_2)}{(M_2\times w_1)}$

$M_2=\frac{(k_b\times 1000\times w_2)}{(\Delta T_b\times w_1)}$

$M_2=\frac{(3.63\times 1000\times 5)}{(3.63\times 20)}=250gmo{l}^{-1}$