The vapour pressure of water at `20^(@)` is `17.5 mmHg`. 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. `17.675mmHg`
B. `15.750mmHg`
C. `16.500mmHg`
D. `17.325mmHg`

Answer

D

Explanation

`"Moles of glucose "=(18)/(180)=0.1`
`"Moles of "H_(2)O=(178.2)/(18)=9.9`
`:.` Mole fraction of solute (glucose)
`x_("solute")=(0.1)/(0.1+9.9)=(0.1)/(10)`
According to Raoult's law
`(p^(@)-p_(s))/(p^(@))=x_("solute")=(17.5-p_(s))/(17.5)=(0.1)/(10)`
`SO" "p_(s)~~17.325mm" "Hg`