Question

A solution is made by mixing of methanol and of acetic acid . Calculate the mole fraction of methanol 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 146. g of methanol and 72. g of acetic acid . Calculate the mole fraction of methanol in this solution. Round your answer to 2 significant digits.

Answer: The mole fraction of methanol is 0.79 and that of acetic acid is 0.21

Explanation:

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

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

• For methanol:

Given mass of methanol = 146. g

Molar mass of methanol = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of methanol}=\frac{146g}{32g/mol}=4.56mol$

• For acetic acid:

Given mass of acetic acid = 72. g

Molar mass of acetic acid = 60 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic acid}=\frac{72g}{60g/mol}=1.2mol$

Mole fraction of a substance is given by:

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

For methanol:

$\chi_{\text{(methanol)}}=\frac{n_{\text{(methanol)}}}{n_{\text{(methanol)}}+n_{\text{(acetic acid)}}}$

$\chi_{\text{(methanol)}}=\frac{4.56}{4.56+1.2}\\\\\chi_{\text{(methanol)}}=0.79$

For methanol:

$\chi_{\text{(acetic acid)}}=\frac{n_{\text{(acetic acid)}}}{n_{\text{(methanol)}}+n_{\text{(acetic acid)}}}$

$\chi_{\text{(acetic acid)}}=\frac{1.2}{4.56+1.2}\\\\\chi_{\text{(acetic acid)}}=0.21$

Hence, the mole fraction of methanol is 0.79 and that of acetic acid is 0.21