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

`P_(CCl_4)=52.43torr`

Step-by-step 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)=(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!