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lapo4ka [179]
3 years ago
15

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

ane and of acetyl bromide .
Chemistry
1 answer:
zubka84 [21]3 years ago
8 0

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

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