When 25 g of a non-volatile solute is dissolved in 100g of water, the vapour pressure is lowered by `2.25 xx 10^(-1)mm`. If the vapour pressure of water at `20^(@)C` is mm, what is the molecular weight of the solute?

Options

A. 206
B. 302
C. 350
D. 276

Answer

( c)

Explanation

Weight of solvent `(w_(1)) = 100g`
Weight of non-volatile solute `(w_(2)) = 25g`
Lowering of vapour-pressure `(p_(i)^(0)-p_(1)) = 0.225 mm`
Vapour pressure of pure solvent `(p_(i)^(0)) = 17.5 mm` 
Molecular weight of solvent, `H_(2)O(M_(1))18g`
Let molecular weight of solute `(M_(2))`=xg According to Raoult's law,
`(P_(i)^(0)-P_(1))/(P_(i)^(0))=(w_(2)xxM_(1))/(M_(2)xxw_(1))`
`(0.225)/(17.5) = (25xx18)/(x xx 100)`
`x=(25xx18xx17.5)/(0.225xx100)`
`x=350g`
`therefore ("Molecular weight of solute") = 350 g`