At 298 K, the vapour pressure of a solution of 7.5 g of non-volatile solute in 90 g of water is 2.8 kPa. If 18 g of water is added to this solution vapour pressure becomes solution the vapour pressure becomes 2.81 kPa at same temperature, the molar mass of solute in g `mol^(-1)` is

Options

A. `17.5`
B. `68.2`
C. `71.5`
D. `51.8`

Answer

C

Explanation

Relation between relative lowering of vapour pressure and molecular mass of solute is given by
`(p^(@) - p)/(p^(@)) = (w_(2)//M_(2))/(w_(1)M_(1))`
Given,
Weight of non- volatile solute, `W_(2) = 7.5 g`
Weight of water, `w_(1) = 90 g`
Vapour pressure of solution `= 2.8` k Pa
In first case,
`(p^(@) - 2.8)/(p^(@)) = (7.5//M_(2))/(90//18)`
`(p^(@)-2.8)/(p^(@)) = (1.5)/(M_(2))` ....(i)
In second case,
`(p^(@) - 2.81)/(p^(@)) = (7.5//M_(2))/(108//18) = (1.25)/(M_(2))` ...(ii)
On solving (i) and (ii) we get
`M_(2) = 71.5 g mol^(-1)`