Chemistry - Liquid Solution Question with Solution | TestHub

Mass of solute, $W_2=46$ g

Molar mass of solute, $M_2=46$ g mol${}^{-1}$

Number of moles of solute, $n_2=\frac{W_2}{M_2}=\frac{46}{46}=1$ mol

Mass of solvent (water), $W_1=162$ g

Molar mass of solvent, $M_1=18$ g mol${}^{-1}$

Number of moles of solvent, $n_1=\frac{W_1}{M_1}=\frac{162}{18}=9$ mol

Using the approximate formula for relative lowering of vapour pressure (as per textbook conventions for dilute solutions):

$\frac{\Delta P}{P^0}\approx \frac{n_2}{n_1}$

$\frac{\Delta P}{P^0}=\frac{1}{9}=\mathrm{0.111...}\approx 0.11$

Answer: 0.11