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

Options

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

Answer

B

Explanation

`"Given, "p_(s)=19.8mm,n_(A)=0.1`
`n_(B)=(178.2)/(18)=9.9`
According to Raoult's law,
`(p^(@)-p)/(p^(@))=(n_(A))/(n_(A)+n_(B))`
`rArr(19.8-p)/(19.8)=(0.1)/(9.9+0.1)`
`rArrp=19.602mm`