The vapor pressure of acetone at `20^(@)C` is 185 torr. When `1.2 g` of a non-volatile solute was dissolved in `100 g` of acetone at `20^(@)C`, it vapour pressure was 183 torr. The molor mass `(g mol^(-1))` of solute is:

Options

A. `64`
B. `128`
C. `488`
D. `32`

Answer

1

Explanation

According to Raoult's law, relative lowering of vapour pressure is equal to the mole fraction of the solute.
`(P_("acetone")^(0)-P_("acetone"))/(P_("acetone")^(0))=n_("solute")/(n_("solvent")+n_("solvent"))`
For dilute solutions, `n_("solute") lt lt n_("solvent")`. Thus
`(P_("acetone")^(0)-P_("acetone"))/(P_("acetone")^(0))=n_("solute")/n_("solvent")`
Let `M (g mol^(-1))` be the molar mass of solute, then
`(1.85 "torr"-183 "torr")/(185 "torr") =(1.2 g//Mg mol^(-1))/(100 g//58 g mol^(-1))`
`2/185=(1.2xx58)/(Mxx100)`
or `M=((1.2)(58)(185))/((100)(2))`
`=64.38 g mol^(-1)`