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

Step-by-step 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}=(49g)/(119g/mol)=0.41moles`

b) moles of
`CH_3COBr`

`\text{Number of moles}=(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}}=(0.41)/(0.41+0.59)=0.41`

The mole fraction of chloroform in this solution is 0.41