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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According to Raoult's law,$$p1=x1p1$$∗$$=1$$−$$x2p1$$∗  i.e.  $$x2=1$$−$$p1p1$$∗$$=1$$−$$183$$ Torr $$185$$ Torr $$=2185Also$$,    $$x2=n2n1+n2=(1.2g/M)100g/58gmol$$−$$1+(1.2g/M)Hence$$,   $$(1.2g/M)100g/58gmol$$−$$1+(1.2g/M)=2185This$$ $$gives1.2gM1$$−$$2185=2185100g58gmol$$−$$1$$  or  $$M=(1.2g)183185185258gmol$$−$$1100g=63.68gmol$$−$$1Approixmate$$ value Writing $$x2$$≃$$n2/n1$$, we get $$2185=1.2g/M100g/58gmol$$−$$1$$. This gives $$M=1.2$$×$$58100$$×$$1852gmol$$−$$1=64.38gmol$$−$$1$$

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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