Vapour pressure of pure water at 298 K is 24.8 mm Hg. Calculate the lowering in vapour pressure of an aqueous solution which freezes at -0 · 3^° C.(…

$\displaystyle \Delta T_f = K_f\, m$$\displaystyle m = \Delta T_f / K_f$$\displaystyle m = 0.3/1.86$$\displaystyle = 0.16\ m$$\displaystyle m = \dfrac{x_2 \times 1000}{M_A}$$\displaystyle x_2 = \dfrac{0.16 \times 18}{1000} = 2.88 \times 10^{-3}$$\displaystyle \dfrac{p_1^{0} - p_1}{p_1^{0}} = x_2$$\displaystyle \dfrac{24.8 - p_1}{24.8} = 2.88 \times 10^{-3}$$\displaystyle p_1^{0} - p_1 = x_2\, p_1^{0}$$\displaystyle = 2.88 \times 10^{-3} \times 24.8\ \text{mm Hg}$$\displaystyle = 0.07\ \text{mm Hg}$