An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at `100^(@)C.` What is the molar mass of the solute?

Answer

`41.35`

Explanation

Relative lowering in vapour pressure.
`( p _(A) ^(@) - p _(S))/( p _(A) ^(@)) - ( n _(B))/( n _(A)) = ( W _(B))/( M _(B)) xx ( M _(A))/(W _(A))`
`p _(A) ^(@)` (water ) = 1.013 bar, `W _(B) = 2 g , M _(A) -= 18 g mol ^(-1)`
`P _(S)` (water)=
`=1. 004` bar,` W _(A) = (100 -2) = 98 g, M _(B) = ?`
Placing the values in equation (i)
`M _(B) = (( 2 g ) xx ( 18 g mol ^(-1)) xx ( 1. 013 "bar"))/( ( 0. 009 "bar") xx ( 98) g)= 41.35 g mol ^(-1)`