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. 17.675 mmHg
B. `15.750 mmHg`
C. `16.500 mmHg`
D. `17.325 mmHg`

Answer

4

Explanation

According to Raoult's law
`(P^(0)-P)/P= chi_("solute")=n_("solute")/(n_("solute")+n_("solvent"))`
`n_("solute")=m_("glucose")/(mm_("glucose"))=(18 g)/(180 g mol^(-1))`
`=0.1 mol`
`n_("solvent")=m_("glucose")/(mm_("glucose"))=(178.2 g)/(180 g mol^(-1))`
`=9.9 mol`
Thus
`(17.5 mmHg-P)/(17.5 mmHg)=0.1/(0.1+9.9)`
Solvent we get
`P=17.325 mmHg`