‘W’ g of a non-volatile electrolyte solid solute of molar mass ‘M’ g , when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. [2026]

Question

‘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) [2026]

Smart Achievers · Accepted Answer

(4)

$P ° = 640 mm Hg$
$P_{s} = 600 mm Hg$
$Δ P = 40 mm Hg$

$moles of solute = \frac{W}{M}$

$\frac{Δ P}{P °} = i \cdot X_{solute}$

$Again:$

$Δ T_{b} = i \cdot K_{b} \cdot m$

$2 = i \times 0.52 \times \frac{W / M}{100} \times 1000$

$i = \frac{2}{5.2} \times \frac{M}{W}$

$X_{solute} = \frac{40}{640} \times \frac{1}{i} = \frac{1}{16} \times \frac{5.2}{2} \times \frac{W}{M}$

$X_{solute} = \frac{1.3}{8} \times \frac{W}{M}$

Solution

1 $\frac{16}{2.6} \times \frac{W}{M}$

2 $\frac{1.3}{8} \times \frac{M}{W}$

3 $\frac{2.6}{16} \times \frac{M}{W}$

4 $\frac{1.3}{8} \times \frac{W}{M}$

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Solution

1 162.6×WM

2 1.38×MW

3 2.616×MW

4 1.38×WM

Ans. (4) P°=640 mm Hg Ps=600 mm Hg ΔP=40 mm Hg moles of solute=WM ΔPP°=i·Xsolute Again: ΔTb=i·Kb·m 2=i×0.52×W/M100×1000 i=25.2×MW Xsolute=40640×1i=116×5.22×WM Xsolute=1.38×WM

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1 $\frac{16}{2.6} \times \frac{W}{M}$

2 $\frac{1.3}{8} \times \frac{M}{W}$

3 $\frac{2.6}{16} \times \frac{M}{W}$

4 $\frac{1.3}{8} \times \frac{W}{M}$