Question

A solution is made by mixing of acetyl bromide and of heptane . Calculate the mole fraction of acetyl bromide in this solution. Be sure your answer has the correct number of significant digits.

Answer 1

The question is incomplete, here is the complete question:

A solution is made by mixing 127.g of acetyl bromide $CH_3COBr$ and 90.g of heptane $C_7H_{16}$. Calculate the mole fraction of acetyl bromide in this solution. Round your answer to 3 significant digits.

Answer: The mole fraction of acetyl bromide is 0.534 and heptane is 0.466

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For acetyl bromide:

Given mass of acetyl bromide = 127. g

Molar mass of acetyl bromide = 123 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetyl bromide}=\frac{127g}{123g/mol}=1.032mol$

• For heptane:

Given mass of heptane = 90. g

Molar mass of heptane = 100.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of heptane}=\frac{90g}{100.2g/mol}=0.90mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For acetyl bromide:

$\chi_{\text{(acetyl bromide)}}=\frac{n_{\text{(acetyl bromide)}}}{n_{\text{(acetyl bromide)}}+n_{\text{(heptane)}}}$

$\chi_{\text{(acetyl bromide)}}=\frac{1.032}{1.032+0.90}\\\\\chi_{\text{(acetyl bromide)}}=0.534$

For heptane:

$\chi_{\text{(heptane)}}=\frac{n_{\text{(acetyl bromide)}}}{n_{\text{(acetyl bromide)}}+n_{\text{(heptane)}}}$

$\chi_{\text{(heptane)}}=\frac{0.90}{1.032+0.90}\\\\\chi_{\text{(heptane)}}=0.466$

Hence, the mole fraction of acetyl bromide is 0.534 and heptane is 0.466