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

Answer:

The answer is "0.35".

Explanation:

please find the complete question in the attached file.

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

Molar mass of $CH_3C0_2 H$ $= 60.05 \ \frac{g}{mol}$

No of moles in $n_{CH_3CO_2H}$ $= 77 \ g \times \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}$  

                                          $= 1.28 \ mol \ CH_3CO_2H$

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

The molar mass of water $(H_2O)$$= 18.02 \ \frac{g}{mol}$

moles of water $n_{H_2O}$:

$= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O$

Molfraction of acetic acid:

$X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}$

                     $=\frac{1.28 \ mol}{1.28 \ mo1 + 2.33 mol}\\\\ = 0.354\\\\= 0.35$