A solution is made by mixing $150g$ of acetyl bromide $(C{H}_3$COBr$)$ and $90\mathrm{.}5g$ of chloroform $(CHC{l}_3).$

a$.$Compute for the mole fraction of acetyl bromide.

ANSWER: The mole fraction of acetyl bromide in the solution is Blank $5.$

b$.$Compute the ppb of chloroform.

ANSWER: The concentration of chloroform is Blank $6{10}^{8}ppb.$