18 g glucose `(C_(6)H_(12)O_(6))` is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is :

Options

A. `759.0`
B. `7.6`
C. `76.0`
D. `752.4`

Answer

D

Explanation

Assuming temperature to be `100^(@)C`
Relative lowering of vapour pressure
Equation `(P^(@)-P^(s))/(P^(@)) = X_("solute") = (n)/(n+N)`
Modified forms of equation is `(P^(@)-P_(s))/(P_(s)) = (n)/(N)`
`P^(@) = 760` torr
`P_(S) = ?`
`(760 -P_(s))/(P_(s)) = (18//180)/(178.2//18)`
`P_(S) = 752.4` torr