Question

What is the vapor pressure of a 45.0 % solution of glucose, c6h12o6, at 90.0°c, given that the vapor pressure of pure water at that temperature is 526 mm hg?

Answer 1

weight of solute = 45 g and weight of solvent = 55 g
M.wt of solute = 180 P⁰ = 526 mmHg
number of moles of solute = 45 / 180 = 0.25 mol
number of moles of solvent = 55 / 18 = 3.05 mol
According to Rault's law:

`(P^(0) - P^(s) )/(P^(0) ) = (n_(solute) )/(n_(solute) + n_(solvent))` = 0.25 / (0.25 + 3.05)


`(526 - P^(s) )/(526) = 0.076`


`P^(s) = 486.15 mmHg`