Multiple choice

A mixture of two immiscible liquids nitrobenzene and water boiling at $97^{o}C$ has a partial vapor pressure of water 733 mm and that of nitrobenzene 27 mm. The ratio of the weights of water to nitrobenzene in the distillate is:

  1. 4

  2. 5

  3. 3

  4. 7
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A Correct answer
Explanation

The ratio of weights of two immiscible liquids in distillate is (P1 * M1) / (P2 * M2). P_water = 733, P_nitro = 27. M_water = 18, M_nitro = 123. Ratio = (733 * 18) / (27 * 123) = 13194 / 3321 = 3.97, which is approximately 4.

AI explanation

For a mixture of immiscible liquids, the ratio of their weights in the distillate is proportional to the product of their vapour pressures and molecular weights. Using the molecular weights of 18 for water and 123 for nitrobenzene gives the ratio as (733 multiplied by 18) divided by (27 multiplied by 123). This calculation results in a value of 13194 divided by 3321, which simplifies to approximately 3.97; rounding to the nearest integer gives the ratio of weights as 4.

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