If the relative lowering in vapour pressure for he ethanolic solution containing non volatile solute is $0.02$ , then the molality of solution will be.(molecular mass of ethanol = $46$ gm/mole)

0.22 m

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1.33 m

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1.09 m

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0.44 m

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Solution

The correct option is D $0.44 m$
By Roulits law
Relative lowering vapour pressure= mole fraction of solute
$\frac{P - P (s)}{P} = X_{2}$

$\frac{[P - P (S)]}{P} = \frac{W \times M}{W \times M} = 0.02$

$\frac{W}{m} \times W = \frac{0.02}{46} = 4.347 \times 10^{- 4}$

$M o l a l i t y = 4.347 \times 10^{- 4} \times 1000 = 0.434 M$

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If the relative lowering in vapour pressure for he ethanolic solution containing non volatile solute is 0.02, then the molality of solution will be.(molecular mass of ethanol = 46 gm/mole)

The correct option is D 0.44mBy Roulits lawRelative lowering vapour pressure= mole fraction of soluteP−P(s)P=X2 [P−P(S)]P=W×MW×M=0.02 Wm×W=0.0246=4.347×10−4Molality=4.347×10−4×1000=0.434M

If the relative lowering in vapour pressure for he ethanolic solution containing non volatile solute is $0.02$ , then the molality of solution will be.(molecular mass of ethanol = $46$ gm/mole)