Question

At a certain temperature the vapor pressure of pure chloroform (CHCl3) is measured to be 91. torr. Suppose a solution is prepared by mixing 140. g of chloroform and 67.1 g of heptane (C,H16) of chloroform and 67.1 g of heptane (C7H16 Calculate the partial pressure of chloroform vapor above this solution.

Answer 1

Answer:

$P_{CCl_4}=52.43torr$

Explanation:

Hello there!

In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

$P_{CCl_4}=x_{CCl_4}P_{CCl_4}^{vap}$

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

$x_{CCl_4}=\frac{140/153.81}{140/153.81+67.1/100.21}=0.576$

Therefore, the partial pressure of chloroform turns out to be:

$P_{CCl_4}=0.576*91torr\\\\P_{CCl_4}=52.43torr$

Regards!