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.

Question

asked Jun 12, 2021 4.1k views

1 Answer

Johannix · Answer 1 · 2021-06-16T11:07:13+0000

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.