Question

QUESTION 2.17

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer 1

Molality of solution is 1 m. It means that one mole of solute is dissolved in 1000 g of water.
Number of moles of solute $\left(n_B\right)=1mol$
Number of moles of water
$\left(n_A\right)=\frac{Massofwater}{Molarmass}=\frac{\left(1000g\right)}{\left(18gmo{l}^{-1}\right)}$
$=55.55mol$
Mole fraction of solute
$\left(x_B\right)=\frac{n_B}{n_A+n_B}=\frac{\left(1mol\right)}{(55.55+1.0)mol}=\frac{1}{56.55}$
$=0.0177$
Mole fraction water $\left(x_A\right)=1-0.0177=0.9823$
Vapour pressure of solution
$\left(p_A\right)=p_A^{\circ }{x}_A=\left(12.3kPa\right)\times 0.9823=12.08kPa$