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

To calculate the mole fraction of acetyl bromide in the solution, divide the moles of acetyl bromide by the total moles of both acetyl bromide and chloroform.

Step-by-step explanation:

To calculate the mole fraction of acetyl bromide in the solution, we first need to calculate the number of moles of acetyl bromide and chloroform. The number of moles of acetyl bromide can be calculated by dividing its mass (124 g) by its molar mass, and the number of moles of chloroform can be calculated using the same method. Then, we can calculate the mole fraction of acetyl bromide by dividing the moles of acetyl bromide by the total moles of both substances. The formula for mole fraction is:

Mole fraction of acetyl bromide = Moles of acetyl bromide / (Moles of acetyl bromide + Moles of chloroform)

Using these values, we can now calculate the mole fraction.

Answer 2

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}=(124g)/(122.95g/mole)=1.008moles`

`\text{Moles of }CHCl_3=\frac{\text{Mass of }CHCl_3}{\text{Molar mass of }CHCl_3}=(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=(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=(0.997)/(1.008+0.997)=0.497`

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