Question

A solution is made by mixing 49.g of chloroform CHCl3 and 73.g of acetyl bromide CH3COBr. Calculate the mole fraction of chloroform in this solution. Round your answer to 2 significant digits.

Answer 1

Answer: The mole fraction of chloroform in this solution is 0.41

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $CHCl_3$

$\text{Number of moles}=\frac{49g}{119g/mol}=0.41moles$

b) moles of $CH_3COBr$

$\text{Number of moles}=\frac{73g}{123g/mol}=0.59moles$

To calculate the mole fraction, we use the formula:

$\text{Mole fraction of a component}=\frac{\text{Moles of the component}}{\text{total moles}}$

$\text{Mole fraction of chloroform}=\frac{\text{Moles of chloroform}}{\text{total moles}}=\frac{0.41}{0.41+0.59}=0.41$

The mole fraction of chloroform in this solution is 0.41