A solution containing 30g of a non-volatile non-electrolyte solute exactly in 90g water has a vapour pressure of `2.8 kPa` at 298K. Further, 18g of water is then added to solution, the new vapour pressure becomes `2.9kPa` at 298K. The solutions obey Raoult's law and are not dilute, molar mass of solute is

Options

A. `23 g mol^(-1)`
B. `34 g mol^(-1)`
C. `43g mol^(-1)`
D. `32 g mol^(-1)`

Answer

A

Explanation

According to Raoult's law
`p_(A) = p_(A)^(@) x_(A)`
`A -=` Water
Let no of moles of solute = n
`2.8 = p_(A)^(@) (5)/(5+n)`
`2.9 = p_(A)^(@) = (5)/(6+n)`
Dividing `(28)/(29)= ((5)xx(6+n))/((5+n)xx(6))`
`(28 xx 6)/(29 xx 5) = (6+n)/(5+n)`
`n = (30)/(23)`
`:.` Molar mass of solute `= 23 g mol^(-1)`
`( :'` Mass of solute = 30g)