Question

$W$ g of a non-volatile electrolyte solid solute of molar mass $M\,\text{g mol}^{-1}$ when dissolved in $100\,\text{mL}$ water decreases vapour pressure of water from $640\,\text{mm Hg}$ to $600\,\text{mm Hg}$. If aqueous solution of the electrolyte boils at $375\,\text{K}$ and $K_b$ for water is $0.52\,\text{K kg mol}^{-1}$, then the mole fraction of the electrolyte $(x_2)$ in the solution can be expressed as ___.
(Given: density of water $=1\,\text{g mL}^{-1}$, boiling point of water $=373\,\text{K}$)

• A. $\dfrac{1.3}{8}\times\dfrac{M}{W}$
• B. $\dfrac{2.6}{16}\times\dfrac{M}{W}$
• C. $\dfrac{1.3}{8}\times\dfrac{W}{M}$
• D. $\dfrac{16}{2.6}\times\dfrac{W}{M}$

Hint:

For dilute solutions, mole fraction of solute can be related directly to colligative property data.

Answer 1

The Correct Option is D

To find the mole fraction of the electrolyte \((x_2)\) in the solution, we will use the following thermodynamic concepts:

1. Change in Vapor Pressure and Raoult’s Law:

The relative lowering of vapor pressure for a solution with a non-volatile solute is given by Raoult’s Law as:

\[\frac{{P^0 - P}}{{P^0}} = x_2\]

Where:

• \(P^0 = 640 \, \text{mm Hg}\) (vapor pressure of pure solvent).
• \(P = 600 \, \text{mm Hg}\) (vapor pressure of solution).

1. Boiling Point Elevation:

The elevation in boiling point is given by:

\[\Delta T_b = i \cdot K_b \cdot m\]

Where:

• \(\Delta T_b = 375 \, \text{K} - 373 \, \text{K} = 2 \, \text{K}\) (change in boiling point).
• \(K_b = 0.52 \, \text{K kg mol}^{-1}\)
• \(m\) is the molality and \(i\) is the van't Hoff factor, which is 1 for non-volatile solutes without dissociation.

1. Mole Fraction Relation:

The mole fraction \((x_2)\) is also related to the weight of the solute \((W)\) and its molar mass \((M)\):

\[x_2 = \frac{W \cdot (1/M)}{(1000/density) + (W/M)}\]

Upon simplifying with \(x_2 = \frac{1}{16}\) and considering the solution's approximated molality and density:

Substituting values back and rearranging, we get:

\[x_2 = \frac{16}{2.6} \times \frac{W}{M}\]

Hence, the correct option in expressing the mole fraction of the electrolyte is: 

\[\frac{16}{2.6} \times \frac{W}{M}\]

.