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Solutions

Question
CBSEENCH12010630

The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1 ) of the substance is:

  • 32

  • 64

  • 128

  • 488

Solution

B.

64

Given, 
po = 185 torr at 20oC
ps = 183 torr at 20oC
Mass of non-volatile substance,
m= 1.2 g
Mass of acetone taken = 100 g
As we have, fraction numerator straight p subscript straight o minus straight p subscript straight s over denominator straight p subscript straight s end fraction space equals space straight n over straight N
putting the values, we get

fraction numerator 185 minus 183 over denominator 183 end fraction space equals space fraction numerator begin display style fraction numerator 1.2 over denominator straight M end fraction end style over denominator begin display style 100 over 58 end style end fraction space
rightwards double arrow 2 over 183 space equals space fraction numerator 1.2 space straight x space 58 over denominator 100 space straight x space straight M end fraction
therefore space straight M space equals space fraction numerator 183 space straight x space 12 space straight x space 58 over denominator 2 space straight x space 100 end fraction
straight M space equals space 63.684 space almost equal to space 64 space straight g divided by mol