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

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:

$\text{Moles of heptane}=\frac{51g}{100.2g/mol}=0.509mol$

• 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:

$\text{Moles of acetyl bromide}=\frac{127g}{123g/mol}=1.032mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of heptane = 0.509 moles

Total moles = [0.509 + 1.032] = 1.541 moles

Putting values in above equation, we get:

$\chi_{(heptane)}=\frac{0.509}{1.541}=0.330$

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