Question

At 25 °C, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous dilute solution of urea is 22.98mm Hg. Calculate the molality of this solution.​

Answer 1

At 25°C, urea is in the form of a non volatile solute.

∴ According to Raoult's law for non volatile solute;

$\frac{P_s}{P^o} = X_1$

⇒ $\frac{22.98}{23.76} = 0.9675$

Now, we know that,

The mole fraction of water = 0.9875

∴ the mole fraction of urea = (1 - 0.9875) = 0.0324

Thus, the Molality of this solution can be calculated as,

$\frac{Number- of -moles-of-urea}{mass-of-water-in-kg}$

= $\frac{0.0324}{0.9875*18} * 1000$ m

= 1.82 m

Ans) The molality of the given solution under the specified temperature is 1.82 Mol

Answer 2

molality is 0.968m under following conditions.