Question

A solution is made by mixing of 42.g water and 77.g of acetic acid HCH3CO2. Calculate the mole fraction of water in this solution.

Answer 1

Moles of $H_2O$ ,

$n_{H_2O}=\dfrac{42}{2\times 1 + 16}=\dfrac{42}{18}\\\\n_{H_2O}=2.33\ moles$

Moles of acetic acid $HCH_3CO_2$ ,

$n_{A.A}=\dfrac{77}{1 + 12 + 3 + 12 + 16\times 2}=\dfrac{77}{60}\\\\n_{A.A}=1.28\ moles$

Mole fraction of water :

$M.F_{H_2O}=\dfrac{n_{H_2O}}{n_{H_2O}+n_{A.A}}\\\\M.F_{H_2O}=\dfrac{2.33}{2.33+1.28}\\\\M.F_{H_2O}=0.645$

Therefore, mole fraction of water in this solution is 0.645 .

Hence, this is the required solution.