Question

If 6 g. of the urea dissolved in $90$ g. of boiling water. The vapour pressure of the solution is:

Answer 1

Rault’s law states that the vapor pressure of a solution of a non-volatile solute is equivalent to the vapor pressure of the pure solvent at that temperature multiplied by its mole fraction. It can be expressed by the following formula:
Vapour pressure $P = P^\circ x,P^\circ =$ Boiling point of pure solvent and x being the mole fraction of the solvent.

Complete step by step answer:
We know,
${P^ \circ } = 760$mm of Hg
Mole fraction of solvent = $\dfrac{\text{moles of solvent}} {\text{total moles of solute and solvent}}$

To find the number of moles of urea (solute) and water (solvent):

Number of moles = $\dfrac{\text{given mass}} {\text{molar mass}}$

Given the mass of urea $= 6g$
Molar mass of urea $= 60g$
Moles of urea( ${n_1}$ )

$= \dfrac{6}{{60}}$

$= 0.1mole$
Given mass of water $= 90g$
Molar mass of water $= 18g$
Moles of water( ${n_2}$ )

$= \dfrac{{90}}{{18}}$
$= 5moles$

Now, to find the vapour pressure of solution we apply the formula:
$P = P^\circ \dfrac{{{n_1}}}{{{n_1} + {n_2}}}$

Substituting values in the formula:

$= 760\left( {\dfrac{{0.1}}{{0.1 + 5}}} \right)$

$= 760 \times 0.0196$

$= 14.89Torr$

Now, $1Torr = 1mmHg$

Vapour pressure of solution $= 760 - 14.89 \approx 744.8$ mm of Hg

Note: Raoult’s law is quite similar to the ideal gas law. The only exception of Raoult’s law is that it applies to solutions too. The ideal gas law assumes the ideal behavior of gases in which the intermolecular forces that are present between dissimilar molecules is zero or non-existent. Whereas, Raoult’s law assumes that the intermolecular forces that exist between different molecules and similar molecules are equal.
Raoult’s law can be applied to non-ideal solutions too. However, this is done by incorporating several factors where we have to consider the interactions between molecules of different substances. An ideal solution is defined as one which obeys Raoult's Law.