The vapour pressure of water at `20^(@)C` is 17.5 mm Hg.
If 18 g of glucose `(C _(6) H _(12) O _(6))` is added to `178. 2 ` g of water at `20 ^(@)C,` the vapour pressure of the resulting solution will be

Options

A. 17. 675 mm Hg
B. 15.750 mm Hg
C. 16.500 mm Hg
D. 17.325 mm Hg

Answer

D

Explanation

Moles of glucose `= (18)/(180) = 0.1 `
Moles of `H _(2) O = ( 178. 2)/(18) = 9.9`
According to Raoult.s law
`( P ^(@) - P _(S))/( P ^(@)) = X _("solute")`
`( 17. 5 - P _(S))/( 17. 5) = (0.1)/(10)`
so,` P _(s) = 17. 325 ` mm Hg