$W \text{ gm}$ of non-volatile electrolyte solute is added in $100 \text{ ml}$ pure water ($\text{P}^\circ = 640 \text{ mm Hg}$) showing vapour pressure of solution $600 \text{ mm Hg}$. This solution have $\text{b}.\text{p}$. of $375 \text{ K}$. Given $K_b$ of $\text{H}_2\text{O} = 0.52 \frac{K \cdot \text{kg}}{\text{mol}}$. Molar mass of solute $= \text{M}$. Select the correct option about mole fraction of solute ($\text{X}_{\text{solute}}$).
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When combining data from different colligative properties, first derive the value of $W/M$ from the most reliable equation (here $\Delta T_b$), and then check which option provides the correct $X_{\text{solute}}$ derived from $\text{RLVP}$.
To solve this question, we need to determine the mole fraction of the solute ($\text{X}_{\text{solute}}$) using the given data and applying relevant formulas. Here's a step-by-step explanation:
First, we use the concept of Raoult’s law. The lowering of vapor pressure of the solvent is given by:
According to the given options, the mole fraction is also expressed in terms of $\frac{W}{M}$, where \(W\) is the weight of the solute and \(M\) is the molar mass of the solute.
The boiling point elevation formula is:
\[\Delta T_b = K_b \times \text{m}\]
Given the boiling point elevation (\(\Delta T_b\)): \(375 \, \text{K} - 373 \, \text{K} = 2 \, \text{K}\)
Using the approximation in dilute solutions where mole fraction \(\text{X}_{\text{solute}}\) can be related to molality and considering the volume of water (approximately 100 g or 0.1 kg):
