Moles of [tex]H_2O[/tex] ,
[tex]n_{H_2O}=\dfrac{42}{2\times 1 + 16}=\dfrac{42}{18}\\\\n_{H_2O}=2.33\ moles[/tex]
Moles of acetic acid [tex]HCH_3CO_2[/tex] ,
[tex]n_{A.A}=\dfrac{77}{1 + 12 + 3 + 12 + 16\times 2}=\dfrac{77}{60}\\\\n_{A.A}=1.28\ moles[/tex]
Mole fraction of water :
[tex]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[/tex]
Therefore, mole fraction of water in this solution is 0.645 .
Hence, this is the required solution.
