Dry air was passed successively through a solution of 5 g of a solutte in 80 g of water and then through pure water. The loss in mass of solution was 2.5 g and that of pure solvent was 0.04 g The molecular mass of the solute is :

Options

A. 48
B. 32
C. 40
D. 35

Answer

A

Explanation

Loss in weight of soluiton `prop p_(A)^(@) - p_(A) = 0.06`
Loss in weight of solvent (water ) `prop p_(A)^(@) - p_(A) = 0.06`
`p_(A)^(@) = (p_(A)^(@) - p_(A)) + p_(A) = 0.06 + 2.5 = 2.56 g`
`(p_(A)^(@) - p_(A))/(p_(A)^(@)) = (w_(B) xx M_(A))/(M_(B) xx w_(B))`
`(0.06)/(2.56) = ( 5 xx 18)/(M_(B) xx 80)`
`M_(B) = ( 5 xx 18 xx 2.56)/(80 xx 0.06) = 48g mol^(-1)`