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The vapour pressure of acetone at 20 °C is 185 Torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 °C, its vapour pressure was 193 Torr. The molar mass of the substance is

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a
32gmol1
b
64gmol1
c
128gmol1
d
488gmol1

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detailed solution

Correct option is B

According to Raoult's law,

p1=x1p1=1x2p1  i.e.  x2=1p1p1=1183 Torr 185 Torr =2185Also,    x2=n2n1+n2=(1.2g/M)100g/58gmol1+(1.2g/M)

Hence,   (1.2g/M)100g/58gmol1+(1.2g/M)=2185

This gives

1.2gM12185=2185100g58gmol1  or  M=(1.2g)183185185258gmol1100g=63.68gmol1

Approixmate value Writing x2n2/n1, we get 2185=1.2g/M100g/58gmol1

This gives M=1.2×58100×1852gmol1=64.38gmol1

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