The relative lowering of vapor pressure of an aqueous solution containing non-volatile solute is $0.0125$ . The molality of the solution is ?

0.70

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0.50

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0.60

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0.0125

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Solution

The correct option is D $0.70$
Relative lowering in vapour pressure $= \frac{W \times M}{M \times W \times 1000}$
Where $W =$ weight of water in kg
$M =$ Molar mass of water in g $= 18 g$
$∴ \frac{W}{M \times N} = M o l a r i t y = \frac{0.0125 \times 1000}{18} = 0.70$

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The relative lowering of vapor pressure of an aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is ?

The correct option is D 0.70Relative lowering in vapour pressure =W×MM×W×1000Where W= weight of water in kgM= Molar mass of water in g=18g∴WM×N=Molarity=0.0125×100018=0.70

The relative lowering of vapor pressure of an aqueous solution containing non-volatile solute is $0.0125$ . The molality of the solution is ?

The correct option is D $0.70$
Relative lowering in vapour pressure $= \frac{W \times M}{M \times W \times 1000}$
Where $W =$ weight of water in kg
$M =$ Molar mass of water in g $= 18 g$
$∴ \frac{W}{M \times N} = M o l a r i t y = \frac{0.0125 \times 1000}{18} = 0.70$