Calculate the reduced vapour pressure of chloroform, CHCl3, at 20 ºC after the a

Question

Calculate the reduced vapour pressure of chloroform, CHCl3, at 20 ºC after the addition of 0.515 g of the non-volatile solute naphthalene, C10H8, to 60.8 g of chloroform. (The vapour pressure of CHCl3 = 156 mmHg at 20 ºC.)
Calculate the reduced vapour pressure of chloroform, CHCl3, at 20 ºC after the addition of 0.515 g of the non-volatile solute naphthalene, C10H8, to 60.8 g of chloroform. (The vapour pressure of CHCl3 = 156 mmHg at 20 ºC.)
Calculate the reduced vapour pressure of chloroform, CHCl3, at 20 ºC after the addition of 0.515 g of the non-volatile solute naphthalene, C10H8, to 60.8 g of chloroform. (The vapour pressure of CHCl3 = 156 mmHg at 20 ºC.)

Explanation / Answer

Calculate the reduced vapour pressure of chloroform, CHCl3, at 20 ºC after the addition of 0.515 g of the non-volatile solute naphthalene, C10H8, to 60.8 g of chloroform. (The vapour pressure of CHCl3 = 156 mmHg at 20 ºC.)

Answer

Given:

Grams of CHCl3 = 60.8 g

Grams of C10H8 = 0.515 g

Vapor pressure of CHCl3 = 156 mmHg at 20 ºC.

First, we need to calculate the moles of naphthalene and moles of chloroform:

Napthalene

0.515 g C10H8 * (1 mol C10H8 /128.17 g C10H8) = 0.004018 mol C10H8

Chloroform

60.8 g CHCl3 * (1 mol CHCl3/119.38 g CHCl3) = 0.50929 mol CHCl3

The total number of moles is 0.004018 + 0.50929 = 0.5133 mol, and the mole fraction of

chloroform is

Mol fraction CHCl3 = 0.51929 mol CHCl3/0.5133 mol = 0.9921

Mol fraction of C10H8 = 0.04018 mol C10H8/0.5133 mol = 0.007828

The vapor-pressure lowering is

P = P° XC10H8 = (156 mmHg)(0.007828) = 1.221 = 1.22 mmHg

Using Raoult's law to calculate the vapor-pressure of chloroform:

P = P°XCHCl3 = (156 mmHg)(0.9921) = 154.7 mmHg

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