2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25ºC, The solution showed a boiling point elevation by 2ºC. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is ________ mm of Hg (Nearest integer)

(Given : Molal boiling point elevation constant of water (K_b) = 0.52 K. kg mol^–1 , 1 atm pressure = 760 mm of Hg, molar mass of water = 18 g mol^–1 ]

The correct answer is 707.

To find the vapour pressure of the resulting aqueous solution, we can use the relationship between the elevation in boiling point and the relative lowering of vapour pressure. We assume the vapour pressure of pure water is 760 mm of Hg (at its boiling point).

Step 1: Calculate the molality (m) of the solution
The elevation in boiling point is given by the formula:
$\Delta T_b=K_b\times m$

Where:
• $\Delta T_b=2\text{K}$ (boiling point elevation)
• $K_b=0.52{\text{K kg mol}}^{-1}$ (molal boiling point elevation constant)

Rearranging the formula to solve for molality ($m$):
$m=\frac{\Delta T_b}{K_b}=\frac{2}{0.52}\approx 3.846{\text{mol kg}}^{-1}$

Step 2: Determine the number of moles of solute and solvent
The mass of the solvent (water), $w_1=100\text{g}=0.1\text{kg}$.

The number of moles of solute ($n_2$) is:
$n_2=m\times w_1\left(\text{in kg}\right)=3.846\times 0.1=0.3846\text{mol}$

The molar mass of water ($M_1$) is 18 g mol^-1. The number of moles of solvent ($n_1$) in 100 g of water is:
$n_1=\frac{100}{18}\approx 5.556\text{mol}$

Step 3: Calculate the vapour pressure of the solution using Raoult's Law
According to Raoult's Law, the relative lowering of vapour pressure is given by:
$\frac{p^{\circ }-p}{p^{\circ }}=\frac{n_2}{n_1+n_2}$

Given the assumption that the solute concentration is negligible with respect to the solvent concentration ($n_2\ll n_1$), the equation simplifies to:
$\frac{p^{\circ }-p}{p^{\circ }}\approx \frac{n_2}{n_1}$

Where:
• $p^{\circ }=760\text{mm of Hg}$ (vapour pressure of pure water)
• $p$ is the vapour pressure of the solution

Substituting the values:
$\frac{760-p}{760}=\frac{0.3846}{5.556}$

$760-p=760\times 0.06922$

$760-p\approx 52.61\text{mm of Hg}$

$p=760-52.61=707.39\text{mm of Hg}$

Rounding to the nearest integer, we get 707 mm of Hg.