Question

A solution is made by mixing of water and of acetic acid . Calculate the mole fraction of water in this solution. Be sure your answer has the correct number of significant digits.

Answer 1

The answer is "0.35".

Step-by-step explanation:

please find the complete question in the attached file.

Mass of
`CH_3C0_2 H`
`= 77 \ g \\\\`

Molar mass of
`CH_3C0_2 H`
`= 60.05 \ (g)/(mol) \\\\`

No of moles in
`n_(CH_3CO_2H)`
`= 77 \ g * (1 \ mol \ CH_3C0_2H)/(60.05 \ g)`

`= 1.28 \ mol \ CH_3CO_2H`

Mass of water
`(H_2O)`
`= 42 \ g`

The molar mass of water
`(H_2O)`
`= 18.02 \ (g)/(mol)`

moles of water
`n_(H_2O)`:

`= 42 \ g * (1 mol H_2O)/(18.02 \ g) \\\\= 2.33 \ mol \ H_2 O`

Molfraction of acetic acid:

`X CH_2CO_2H = (n_(CH_3CO_2H))/(n_CH_3CO_2H +n_(H_2))\\\\`

`=(1.28 \ mol)/(1.28 \ mo1 + 2.33 mol)\\\\ = 0.354\\\\= 0.35`