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

Explanation :  Given,

Mass of = 124 g

Molar mass of = 122.95 g/mole

Mass of = 119 g

Molar mass of = 119.38 g/mole

First we have to calculate the moles of and .

Now we have to calculate the mole fraction of and .

and,

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

wher is the rest of the question