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Ronch [10]
3 years ago
11

At a certain temperature the vapor pressure of pure chloroform (CHCl3) is measured to be 91. torr. Suppose a solution is prepare

d 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.
Chemistry
1 answer:
OleMash [197]3 years ago
3 0

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!

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(b) n_1=4 to  n_2 = 20

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(c) n_1=3 to  n_2 = 10

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