Chemistry - Liquid Solution Question with Solution | TestHub

Given:
$w_2=10.0$ g
$w_1=0.2$ kg
$T_b=81.2^{\circ }$C
$T_b^{\circ }=80.1^{\circ }$C
$K_b=2.53$ K kg mol${}^{-1}$

Elevation in boiling point is given by:
$\Delta T_b=T_b-T_b^{\circ }=81.2-80.1=1.1$ K

Using the formula for elevation in boiling point:
$\Delta T_b=K_b\times m$
$\Delta T_b=K_b\times \frac{w_2}{M_2\times w_1}$

Substituting the given values:
$1.1=2.53\times \frac{10.0}{M_2\times 0.2}$

$1.1=\frac{2.53\times 50}{M_2}$

$M_2=\frac{126.5}{1.1}=115$ g mol${}^{-1}$

Answer: $115$ gm/mol