Question

A solution is made by mixing 124. g of acetyl bromide (CH COBI and 119. g of chloroform (CHCI3) Calculate the mole fraction of acetyl bromide in this solution. Round your answer to 3 significant digits.

Answer 1

Answer : The mole fraction of acetyl bromide in this solution is, 0.503

Explanation :  Given,

Mass of $C_2H_3BrO$ = 124 g

Molar mass of $C_2H_3BrO$ = 122.95 g/mole

Mass of $CHCl_3$ = 119 g

Molar mass of $CHCl_3$ = 119.38 g/mole

First we have to calculate the moles of $C_2H_3BrO$ and $CHCl_3$.

$\text{Moles of }C_2H_3BrO=\frac{\text{Mass of }C_2H_3BrO}{\text{Molar mass of }C_2H_3BrO}=\frac{124g}{122.95g/mole}=1.008moles$

$\text{Moles of }CHCl_3=\frac{\text{Mass of }CHCl_3}{\text{Molar mass of }CHCl_3}=\frac{119}{119.38g/mole}=0.997moles$

Now we have to calculate the mole fraction of $C_2H_3BrO$ and $CHCl_3$.

$\text{Mole fraction of }C_2H_3BrO=\frac{\text{Moles of }C_2H_3BrO}{\text{Moles of }C_2H_3BrO+\text{Moles of }CHCl_3}$

$\text{Mole fraction of }C_2H_3BrO=\frac{1.008}{1.008+0.997}=0.503$

and,

$\text{Mole fraction of }CHCl_3=\frac{\text{Moles of }CHCl_3}{\text{Moles of }C_2H_3BrO+\text{Moles of }CHCl_3}$

$\text{Mole fraction of }CHCl_3=\frac{0.997}{1.008+0.997}=0.497$

Therefore, the mole fraction of acetyl bromide in this solution is, 0.503