Question # The vapour pressure of acetone at ( 20^{circ} mathrm{C} ) is 185 torr. When ( 1.2 mathrm{g} ) of a non-volatile substance was dissolved in ( 100 mathrm{g} ) of acetone at ( 20^{circ} mathrm{C}, ) its vapour pressure was 183 Torr. The molar mass of the substance is

# The vapour pressure of acetone at ( 20^{circ} mathrm{C} ) is 185 torr. When ( 1.2 mathrm{g} ) of a non-volatile substance was dissolved in ( 100 mathrm{g} ) of acetone at ( 20^{circ} mathrm{C}, ) its vapour pressure was 183 Torr. The molar mass of the substance is

(a) 32

(b) 64

(a) 128

(b) 488

Solution

Given, ( p^{circ}=185 ) Torr at ( 20^{circ} mathrm{C} )

[

p_{s}=183 text { Torr at } 20^{circ} mathrm{C}

]

Mass of non-volatile substance, ( m=1.2 mathrm{g} ) Mass of acetone taken ( =100 mathrm{g} )

[

M=?

]

As, we have ( frac{p^{circ}-p_{s}}{p_{s}}=frac{n}{N} )

Putting the values, we get, ( frac{185-183}{183}=frac{frac{1.2}{11}}{frac{100}{58}} )

( Rightarrow quad frac{2}{183}=frac{1.2 times 58}{100 times M} )

( therefore quad M=frac{183 times 1.2 times 58}{2 times 100} )

( M=63.684=64 mathrm{g} / mathrm{m} Omega )