JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
'W' g of a non-volatile electrolyte solid solute of molar mass 'M' g \( mol^{-1} \) when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and \( K_{b} \) for water is 0.52 K kg \( mol^{-1} \), then the mole fraction of the electrolyte solute \( (X_{2}) \) in the solution can be expressed as
(Given density of water = 1 g/mL and boiling point of water = 373 K)
- A \( \frac{1.3}{8} \times \frac{W}{M} \)
- B \( \frac{16}{2.6} \times \frac{W}{M} \)
- C \( \frac{2.6}{16} \times \frac{M}{W} \)
- D \( \frac{1.3}{8} \times \frac{M}{W} \)
Answer & Solution
Correct Answer
(A) \( \frac{1.3}{8} \times \frac{W}{M} \)
Step-by-step Solution
Detailed explanation
P° = 640 mm Hg \(P_S = 600 mm Hg\) \( \Delta p = 40 \) mm Hg moles of solute = \( \frac{W}{M} \) \( \frac{\Delta P}{P^{\circ}}=i.X_{solute} \) Again: \( \Delta T_{b}=i \times k_{b} \times m \) \( 2=i \times 0.52 \times \frac{W/M}{100} \times 1000 \)…
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