The vapour pressure of water at `20^(@)` C is 17.5 mm Hg. If 18 g of gulucose`(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. `17.325 mm Hg`
B. `17.675 mm Hg`
C. `15.750 mm Hg`
D. `16.500 mm Hg`

Answer

A

Explanation

Mole of glucose ` = (18)/(180) = 0.1`
Mole of water `= (178.2)/(18) = 9.9`
`(p^(@) -p)/(p^(@)) = (0.1)/(0.1+9.9)`
`(17.5 - p)/(17.5) = (0.1)/(10)`
`17.5 - p = (0.1xx17.5)/(10) = 0.175`
` p = 17.5 - 0.175`
` = 17.325 mm Hg`