An aqueous solution is made by dissolving 10g of glucose `(C_(6)H_(12)O_(6))` in 90g of water at 300K. If the vapour pressure of pure water at 300 K is 32.8 mm Hg, what would be the vapour pressure of the solution?

Answer

32.44 mm

Explanation

Weight of glucose = 10 g
Mol. wt. of glucose `(C_(6)H_(12)O_(6))=12xx6+1xx12 +16xx6`
`=180" g mol"^(-1)`
No. of moles of glucose`=(10)/(180)=0.055`,
Weight of water = 90 g , No. of moles of water `=(90)/(18)=5`
Mole fraction of glucose `=(0.055)/(5+0.055)=0.011`
`(p^(0)-p)/(p^(0))=x_("glucose")`
Substituting the values, `(32.8-p)/(32.8)=0.011`
or `32.8-p=32.8xx0.011=0.36`
`:.p=32.8-0.36=32.44` mm