Vapour pressure varing with molality

Question

Vapour pressure of water at 298 K is 0.025bar. Following solutions are prepared at the given temperature whose molality I have calculated to be,

Solution A 0.56m urea dissolved in 178.2g of water
Solution B 0.00556m glucose dissolved in 179.2g of water
Solution C 0.279m of sodium carbonate dissolved in 179.1g of water.

Identify the correct order in which the vapour pressure increases.

Edit: I have calculated the molality of the three solutions, now it simply requires some kind of relation to be established between vapour pressure and molality which I am not getting.

Answer 1

According to Raoult's law, the vapour pressure of a solution depends on the vapor pressure of its components and their mole fraction $$ p = p^{\star}_{\rm A} x_{\rm A} + p^{\star}_{\rm B} x_{\rm B} + \cdots $$ Where

$p^{\star}_{\rm i}$ is the vapor pressure of the pure component $i$

$x_{\rm i}$ is the mole fraction of the component $i$ in the mixture (in the solution).

As urea, glucose and sodium carbonate are non volatile (their partial pressure is zero). The overall vapor pressure depends only on the water's vapor pressure and mole fraction $$ p = p^{\star}_{\rm water} x_{\rm water} $$

Reference: Raoult's law