Question

A solution is made by mixing of 51 g of heptane and of acetyl bromide . Calculate the mole fraction of heptane in this solution. Round your answer to significant digits.

Answer 1

The question is incomplete, here is the complete question:

A solution is made by mixing of 51 g of heptane and 127 g of acetyl bromide. Calculate the mole fraction of heptane in this solution. Round your answer to 3 significant digits.

Answer: The mole fraction of heptane in the solution is 0.330

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For heptane:

Given mass of heptane = 51 g

Molar mass of heptane = 100.2 g/mol

Putting values in equation 1, we get:

`\text{Moles of heptane}=(51g)/(100.2g/mol)=0.509mol`

• For acetyl bromide:

Given mass of acetyl bromide = 127 g

Molar mass of acetyl bromide = 123 g/mol

Putting values in equation 1, we get:

`\text{Moles of acetyl bromide}=(127g)/(123g/mol)=1.032mol`

Mole fraction of a substance is given by:

`\chi_A=(n_A)/(n_A+n_B)`

Moles of heptane = 0.509 moles

Total moles = [0.509 + 1.032] = 1.541 moles

Putting values in above equation, we get:

`\chi_((heptane))=(0.509)/(1.541)=0.330`

Hence, the mole fraction of heptane in the solution is 0.330