Question

At a certain temperature the vapor pressure of pure heptane is measured to be . Suppose a solution is prepared by mixing of heptane and of acetyl bromide .

Answer 1

This is an incomplete question, here is a complete question.

At a certain temperature the vapor pressure of pure heptane is measured to be 170 torr. Suppose a solution is prepared by mixing 86.7 g of heptane and 125 g of acetyl bromide.

Calculate the partial pressure of heptane above this solution. Round your answer to 3 significant digits.

Answer : The partial pressure of heptane is, 78.2 torr

Explanation : Given,

Mass of heptane = 86.7 g

Mass of acetyl bromide = 125 g

Molar mass of heptane = 100 g/mole

Molar mass of acetyl bromide = 122.9 g/mole

First we have to calculate the moles of heptane and acetyl bromide.

$\text{Moles of heptane}=\frac{\text{Mass of heptane}}{\text{Molar mass of heptane}}=\frac{86.7g}{100g/mole}=0.867mole$

and

$\text{Moles of acetyl bromide}=\frac{\text{Mass of acetyl bromide}}{\text{Molar mass of acetyl bromide}}=\frac{125g}{122.9g/mole}=1.017mole$

Now we have to calculate the mole fraction of heptane and acetyl bromide.

$\text{Mole fraction of heptane}=\frac{\text{Moles of heptane}}{\text{Moles of heptane}+\text{Moles of acetyl bromide}}=\frac{0.867}{0.867+1.017}=0.460$

and

$\text{Mole fraction of acetyl bromide}=\frac{\text{Moles of acetyl bromide}}{\text{Moles of heptane}+\text{Moles of acetyl bromide}}=\frac{1.017}{0.867+1.017}=0.539$

Now we have to partial pressure of heptane.

$p_{heptane}=X_{heptane}\times p^o_{heptane}$

where,

$p^o_{heptane}$ = partial pressure of heptane

$p_{heptane}$ = total pressure of gas

$X_{heptane}$ = mole fraction of heptane

$p_{heptane}=X_{heptane}\times p^o_{heptane}$

$p_{heptane}=0.460\times 170torr=78.2torr$

Therefore, the partial pressure of heptane is, 78.2 torr