MEDIUMMCQ SINGLEJEE Advanced ChemistryPhysical ChemistrySolutionsRelative lowering of vapour pressure
The vapour pressure of acetone at 20°C is 185 torr. When 1.2g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is:
A.128
B.488
C.32
D.64✓ Correct
Explanation
The molar mass 2 M of non-volatile substance can be calculated from the following formula. 2 1 1 2 W M p p W M p x x Here, pvapour pressure of acetone and p is the vapour pressure of the solution. 1 M and 1 W are the molar mass and mass of acetone. 2 M and 2 W are the molar mass and mass of a non- volatile substance. Substitute values in the above expression. 2 185 183 1.2 58 185 100 M x x Hence, 2 M 64.38g/mol. Thus, the molar mass of the non-volatile substance is 64 g/mol. Elevation in boiling point
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