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

Options

A. 7.6
B. 76
C. 752.4
D. 759

Answer

C

Explanation

At 1 atm , the vapour pressure of water , `p^(@) = 760 torr`
` w_(B) = 18 g , M_(B) = 180 g mol^(-1) , w_(A) = 178.2 = 2711 g, p_(A) = ?`
`therefore " "(p_(A)^(@) - p_(A))/(p_(A)^(@)) = (w_(B)M_(A))/(w_(A)M_(B))`
or `(760 -p_(A))/(760) = (18 xx 18)/(178.2 xx 180)`
or ` (760 - p_(A))/(760) = 0.0101`
or `p_(A) = 752.3 torr`