Question

What would be the molar mass of a compound if $6.21$ g of it dissolved in $24.0$ g of chloroform form a solution that has a boiling point of ${68.04}^o$C. The boiling point of pure chloroform is ${61.7}^o$C and the boiling point elevation constant, $K_b$ for Chloroform is ${3.63}^o$ c/m.

Answer 1

mass of solute $\left(n_{solute}\right)=6.21g$

mass of chloroform $\left(solvent\right)=24.0g$

Boiling point of the solution $={68.04}^{\circ }C$

Boiling point of the chloroform $={61.70}^{\circ }C$

$\Delta T_b=\left({68.04}^{{}^{\circ }}C-{61.70}^{{}^{\circ }}C\right)={6.34}^{\circ }C$

$K_b={3.63}^{\circ }Cmo{l}^{-1}K{g}^{-1}$

$\Delta =K_b.m$

$=K_b\times \frac{n_{solute}}{massofchloroforminKG}$

${6.34}^{{}^{\circ }}C={3.63}^{{}^{\circ }}Cmo{l}^{-1}Kg\times \frac{6.21\frac{g}{M}}{0.024Kg}=148.1gmo{l}^{-1}$