The vapour pressure of water at `23^(@)C` is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm) of the resultant solution?

Options

A. `19.0`
B. `19.602`
C. `19.402`
D. `19.202`

Answer

(b)

Explanation

Vapour pressure of water `(p^(0i)) - 19.8 mm`
Number of moles of glucose `(n_(1))= 0.1`
Weight of water,`H_(2)O(w) = 178.2g`
Number of moles of `H_(2)O (n_(2))("Weight of " H_(2)O)/("Molecular weight of "H_(2)O) = (178.2)/(18)`
`n_(2) = 9.9`
According to Raoult's law,
`(P_(i)^(0)-p_(i))/(P_(i)^(0)) = (n_(2))/(n_(1))+(n_(2))`
`(19.8-P_(i)^(0))/(19.8)= (9.9)/(0.1+9.9)`
`((19.8-P_(1)))/(19.8)=(0.1)/(10)`
`198-10P_(1)=19.8xx0.1`
`10P_(1)=19.8-1.98`
`10P_(1)=196.02`
`P_(1)=19.602 mm`
`therefore` Vapour pressure of resultant solution = 19,602 mm