For a dilute solution containing $2.5 g m$ of a non-volatile non-electrolyte solute in $100 g m$ of water, the elevation in boiling point at $1 a t m$ pressure is $2^{o} C$ . Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of $H g$ ) of the solution is ( $K_{b} = 0.76 K k g m o l^{- 1})$ :

724

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740

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736

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718

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Solution

The correct option is C $724$
$Δ T_{b} = K_{b} \times m$
$2 = 0.76 \times \frac{2.5 \times 1000}{M o l . w t . o f s o l u t e \times 100}$
$∴$ Mol. wt. of solute $= 9.5 g m o l^{- 1}$
Again,

$\frac{Δ P}{760} = \frac{2.5 / 9.5}{100 / 18}$
Lowering of V.P. $= 36$
$∴$ Vapour pressure of solution $= 760 - 36 = 724$

Option A is correct.

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The correct option is C 724ΔTb=Kb×m2=0.76×2.5×1000Mol.wt. of solute×100∴ Mol. wt. of solute =9.5 g mol−1Again,ΔP760=2.5/9.5100/18Lowering of V.P. =36∴ Vapour pressure of solution =760−36=724Option A is correct.