18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure of the solution is equal to

Options

A. 6
B. 0.2
C. 5.1
D. 0.02

Answer

D

Explanation

From Raoult.s law for non-volatile solute, we know that
`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))`
For dilute solution, it becomes
`(p^(@)-p_(s))/(p^(@))=(n_s)/(n_1)=(w_2)/(m_2)xx(M_1)/(W_1)`
Given that, `w_(2)=18g,W_(1)=90g`
Molecular mass of glucose
`(C_(6)H_(12)O_(6))=12xx6+12xx1+6xx16=180`
Molecular mass of water `(H_(2)O)+2+16=18`
Now, putting the values, we get
`(p^(@)-p_(s))/(p^(@))=(18)/(180)xx(18)/(90)=0.02`