Question

Two liquids A and B are miscible over the whole range of composition and may be treated as ideal. At 350 K the vapour pressure of pure A is 24 kPa and of pure b is 12 kPa .A mixture of 60% A and 40% B is distilled at this temperature. a small amount of the distillate is collected and redistilled at 350K.

what is the mole percent of B in the second distillate ?

$\rule{200}{2}$
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Answer 1

Answer:

$\large \text{Mole \% \ of \ x_B=14.28\% }$

Explanation:

We have find mole percent of B.

Given :

Mixture of 60% A and 40% B is distilled at this temperature.

$\large x_A=0.6 \ and \ x_B=0.4$

$\large Given \ P_A=24 \ kPa \ and \ P_B=12 \ kPa$

Now we have

$\large P_T=P_A\times x_A+P_B\times x_B$

Put the values here we get

$\large \text{ P_T=24\times 0.6+12\times0.4 kPa}\\\\\\\large \text{ P_T=14.4+4.8 kPa}\\\\\\\large \text{ P_T=19.2 kPa}$

We also know that

$\large y_B=\dfrac{P_B}{P_T} \ and \ y_A=\dfrac{P_A}{P_T} \\\\\\\large y_B=\dfrac{4.8}{19.2}=0.75\\\\\\\large y_A=\dfrac{14.4}{19.2}=0.25 \\\\\\\large P_{sol}=24\times0.75+12\times0.25\\\\\\\large P_{sol}=18+3=21\\\\\\\large y_A=\dfrac{6}{7}\\\\\\\large y_B=\dfrac{3}{21}=\dfrac{1}{7}$

$\large \text{Mole \% \ of \ x_B=\dfrac{1}{7}\times100 }\\\\\\\large \text{Mole \% \ of \ x_B=14.28\% }$

Thus we get answer.