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At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va

Bas_tet [7]

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

$P=X_{A}P_{A}+X_{B}P_{B}$

Here, $X_{A}$ is mole fraction of A, $X_{B}$ is mole fraction of B, $P_{A}$ is partial pressure of A and $P_{B}$ is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

$X_{A}+X_{B}=1$

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

$X_{octane}=1-X_{hexane}=1-0.58=0.42$

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

$P=X_{hexane}P_{hexane}+X_{octane}P_{octane}$

Putting the values,

$P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg$

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

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