Question

A solution containing 2 g of glucose (M = 180 g mol⁻¹) in 100 g of water is prepared at 303 K. If the vapour pressure of pure water at 303 K is 32.8 mm Hg, what would be the vapour pressure of the solution? 

Answer 1

`"Number of moles of water" = ("Mass of water")/("mol mass of water")`

= `100/18`

= 5.55 moles

`"Number of moles of glucose" = ("Mass of glucose")/("mol mass of glucose")`

= `2/180`

= 0.0111

Total moles = 5.55 + 0.011

= 5.56 mol

`x_("Water") = 5.55/5.56`

`"P"_("solution") = "x"_("water") × "p"_("water")`

P_{s =} 0.99 × 32.8

= 32.74 mm of Hg