The weight in grams of a non-volatile solute (mol.wt.60) to be dissolved in 90g of water to produce a relative lowering of vapour pressure of 0.02 is

Options

A. 4
B. 8
C. 6
D. 10

Answer

c

Explanation

Relative lowering of vapour pressure,`[(P^(0)-P)/P^(0)]=0.02`
GMW of solute, `M_(A)` = 60
Weight of water, `W_(B)` = 90 g
` therefore (P^(0)-P)/P^(0)=X_(A)=n_(A)/(n_(A)+n_(B))= (W_(A)/M_(A))/( (W_(A)/M_(A))+(W_(B)/M_(B))`
`0.02=(W_(A)/60)/( (W_(A)/60)+(90/18))= (W_(A)/60)/( (W_(A)/60)+5`
`W_(A)/60 +5=(W_(A)/60)xx100/2`
`(50W_(A))/60-(W_(A)/60)=5`
`49W_(A)=300`
`W_(A)=300/49`
`W_(A)=6.122 approx 6`