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)=(42)/(2* 1 + 16)=(42)/(18)\\\\n_(H_2O)=2.33\ moles`

Moles of acetic acid
`HCH_3CO_2` ,

`n_(A.A)=(77)/(1 + 12 + 3 + 12 + 16* 2)=(77)/(60)\\\\n_(A.A)=1.28\ moles`

Mole fraction of water :

`M.F_(H_2O)=(n_(H_2O))/(n_(H_2O)+n_(A.A))\\\\M.F_(H_2O)=(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.