Question

(a) Vapour pressure of chloroform $\left(CHC{l}_3\right)$ and dichloromethane $\left(C{H}_{2}C{l}_2\right)$ at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of $C{H}_{2}C{l}_{2}at298K$

(b) State Raoult’s law

(c) What is meant by azeotropicmixture?

Answer 1

(a) V.P of $CHC{l}_3=200$ mm of Hg

V.P of $C{H}_{2}C{l}_2=415$ mm of Hg

Molar mass of $C{H}_{2}C{l}_2=85$

Molar mass of $CHC{l}_3=119.5$

Moles of $C{H}_{2}C{l}_2=\frac{40}{85}=0.47$

Moles of $CHC{l}_3=\frac{25.5}{119.5}=0.213$

Total moles = 0.47 + 0.213 = 0.683 mol

$X_{C{H}_{2}C{l}_2}=\frac{0.47}{0.683}=0.688X_{C{H}_{2}C{l}_2}$
$X_{CHC{l}_3}=1.000-0.688=0.312$
$P_{total}=P_1^{\circ }+(P_2^{\circ }-P_1^{\circ })X_2=200+(415-200)\times 0.688$
$=200+147.9=347.9mmofHg$

(b) Raoult’s law – For any solution the partial vapour pressure of each volatile compound in the solution is directly proportional to the mole fraction.

(c) Azeotropes: Binary mixtures having the same composition in liquid and vapour phase and boil at constant temperature.