A solution of carbon tetrachloride and acetic acid (CH3COOH) that is 50% carbon tetrachloride by mass is boiling at 98.6 °C. The vapor is collected and cooled until it condenses to form a new solution. Calculate the percent by mass of carbon tetrachloride in the new solution.

The percent by mass of carbon tetrachloride in the new solution is 72.6

Explanation:

Molecular weight of Carbon Tetrachloride is 154 g/mol

Molecular weight of CH3COOH is 60g/mol

Mass fraction of CCl4 is 0.5

Mass fraction of CCl4 is (0.5/154)/{(0.5/154) + (0.5/60)}

Mass fraction of CCl4 is X = 0.2803 and Y = 1-X

As per Raoul’s law

Y Pt = X * P1

Pt = X * P1 + YP2

Pt = 0.2803 * 1377 + (1-0.2803) * 519

Pt = 759.557 torr

Substituting the given values we get  

Y1 = X1P1/Pt

Y1 = 0.2803 * 1377/759.557

Y1 = 0.508

Mass fraction  

(0.508)(154)/{( (0.508)(154)) + (1-0.508)(60)}

= 0.726 = 72.6 %

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