Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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.

Sagot :

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

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]

a) moles of [tex]CHCl_3[/tex]

[tex]\text{Number of moles}=\frac{49g}{119g/mol}=0.41moles[/tex]

b) moles of [tex]CH_3COBr[/tex]

[tex]\text{Number of moles}=\frac{73g}{123g/mol}=0.59moles[/tex]

To calculate the mole fraction, we use the formula:

[tex]\text{Mole fraction of a component}=\frac{\text{Moles of the component}}{\text{total moles}}[/tex]

[tex]\text{Mole fraction of chloroform}=\frac{\text{Moles of chloroform}}{\text{total moles}}=\frac{0.41}{0.41+0.59}=0.41[/tex]

The mole fraction of chloroform in this solution is 0.41