The relative lowering of vapour pressure of an aqueous solution containing a non volatile solute is $0.0125$ . The molality of the solution is:

0.70

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0.30

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0.125

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0.07

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Solution

The correct option is A $0.70$
According to Raoult's law,
Relative lowering of vapour pressure $=$ mole fraction of solute.

$(X_{B}) = 0.0125$

$∵ X_{B} = \frac{m \cdot M_{A}}{1000 + m \cdot M_{A}}$

Where m=molality,

$∴ 0.0125 = \frac{m \times 18}{1000 + m \times 18}$

$12.5 + 0.225 m = 18$ m

$17.775 m = 12.5$

$m = \frac{12.5}{17.775} = 0.70$ .

Hence, option A is correct.

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The relative lowering of vapour pressure of an aqueous solution containing a non volatile solute is 0.0125. The molality of the solution is:

The correct option is A 0.70According to Raoult's law, Relative lowering of vapour pressure = mole fraction of solute.(XB)=0.0125∵XB=m⋅MA1000+m⋅MAWhere m=molality,∴0.0125=m×181000+m×1812.5+0.225 m=18 m17.775 m=12.5m=12.517.775=0.70.Hence, option A is correct.

The relative lowering of vapour pressure of an aqueous solution containing a non volatile solute is $0.0125$ . The molality of the solution is: