Calculate vapour pressure of 360 ml \(\frac{M}{9}\) glucose solution at 294 K. (Given that vapour pressure of water at 294 K is 18 mm Hg and density of solution is 1.2 g/ml)
Answer & explanation
Correct answer: option 3
M = \(\frac{\text{moles of glucose × 1000}}{\text{volume of solution (in ml)}}\)
\(\frac{1}{9}\) = \(\frac{n}{360}\) x 1000
n = 0.04 = moles of glucose
Weight of glucose = 0.04 x 180 = 7.2 g
Weight of glucose = 360 x 1.2 = 432 g
Weight of water = 432 - 7.2 = 424.8 g
Moles of water = \(\frac{424.8}{18}\) = 23.6
Mole fraction of glucose = \(\frac{0.04}{23.6 + 0.04}\) = 0.0017
According to Relative lowering in vapour pressure
\(\frac{Po - p}{Po}\) = 0.0017
Po - P = 0.0017 x 18
18 - P = 0.03
18 - 0.03 = P
P = 17.97 mm Hg.