Chemistry - Liquid Solution Question with Solution | TestHub

Mass of solute, $W_2=0.56$ g

Molar mass of solute, $M_2=60$ g mol${}^{-1}$

Number of moles of solute, $n_2=\frac{W_2}{M_2}=\frac{0.56}{60}=0.00933$ mol

Mass of solvent (water), $W_1=100$ g

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

Number of moles of solvent, $n_1=\frac{W_1}{M_1}=\frac{100}{18}=5.555$ mol

Relative lowering of vapour pressure is equal to the mole fraction of the solute, $x_2$:

$x_2=\frac{n_2}{n_1+n_2}$

$x_2=\frac{0.00933}{5.555+0.00933}=\frac{0.00933}{5.5643}\approx 0.001677$

Rounding off, the relative lowering of vapour pressure is $0.0017$.

Answer: $0.0017$