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# Partial pressure of immiscible liquids

Let ‘A’ and ‘B’ be the two volatile and immiscible liquids; pA and pB be the partial pressures of ‘A’ and ‘B’ respectively.

Then pA/pB = MA/MB

where mA and mB are molar concentrations of ‘A’ and ‘B’ respectively.

pA/pB = WA/mA/WB/mB

Where WA, WB are weights of ‘A’ and ‘B’ and mA, mB are molecular weights of ‘A’ and ‘B’ respectively.

Example:

Phenol associated in benzene to a certain extent for a dimer. A solution containing 20 × 10−3 kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. Calculate the fraction of the phenol that has dimerised. (Kf of benzene is 5.12oK kg mol−1)

Solution:

Observed mol. mass

= 1000×Kf×w/W?T

= 1000×5.12×20×10–3/1×0.69 = 148.4

Normal mol. mass of phenol (C6H5OH) = 94

So Normal mol. mass/Observed mol. mass = 94/148.4

= 1 + (1/n + 1)α = 1 + (1/2 – 1)α

94/148.4 = 1–α/2

or α = 0.733 or 73.3 %

Example:

The molal depression of the freezing point in 1000 g of water is 1.86. Calculate what would be the depression of freeing point of of water when (a) 120 g of urea. (b) 117 g of sodium chloride, (c) 488.74 g of BaCl2.2H2O have been dissolved in 1000 g of water. It is assumed that sodium chloride and barium chloride are fully ionized.

Solution:

(a) Urea is a non-electrolyte. The number of particles does not change in solution.

Thus, ?T = 1000×Kf×w/W×m = 1000×1.86×120/1000×60 = 3.72oC

No. of moles of urea = 120/60 = 2

(b) No. of moles of NaCl = 117/58.5 = 2

NaCl is an electrolyte. It ionizes compeletely.

One molecule of NaCl furnishes 2 ions (one Na+ and one Cl).

Hence, the number of particles will be double as compared to urea solution.

So ?TNaCl = 2 × ?Turea = 2 ×3.72 = 6.44oC

(c) One molecule of BaCl2.2H2O furnishes three ions (one Ba2+ and two Cl)

No. of moles of BaCl2.H2O = 488.74/244.37 = 2

Hence, the number of particles will be three times as compared to the urea solution.

So ?TBaCl2 = 3?Turea = 3 × 3.72 = 11.16oC

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