Discussion Forum

 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 183 torr. The molar mass  (gmol^-1)  of the substance is

Answer:

Explanation:

Given

   P₀   =185 torr at 20° C

   P_s    =183 torr at 20° C

 Mass of non-volatile substance,

                   m=1.2 g

Mass of acetone taken  =100g

              M=  ?

As, we have

           $\frac{P_{0}-P_{s}}{P_{s}}=\frac{n}{N}$

 Putting the values , we get,

  $\frac{185-183}{183}=\frac{\frac{1.2}{M}}{\frac{100}{58}}\Rightarrow\frac{2}{183}=\frac{1.2\times58}{100 \times M}$

$\therefore$                $M=\frac{183\times1.2\times58}{2\times100}$

   M=63.684 =64g/mol