Question

The volume of 1 M aqueous $NaOH$ solution required to completely react with 71 g of an aqueous acetic acid solution in which mole fraction of the acid is 0.18 is $\frac{1}{x}$ L. Calculate the value of x.

• A. 2
• B. 4
• C. 3
• D. 6

Answer 1

The correct option is A 2

Let us assume the weight of acetic acid present in solution be w g.
$NaOH+C{H}_{3}COOH\to C{H}_{3}COONa+H_{2}O$
Mole fraction of acetic acid in solution = $x_{\text{acetic acid}}=0.18$
$x_{\text{acetic acid}}$ = $\frac{n_{\text{acetic acid}}}{n_{\text{acetic acid}}+n_{\text{water}}}$
$x_{\text{acetic acid}}$ = $\frac{w/60}{w/60+(71-w)/18}$ = 0.18
On solving, w = 30 g
moles of acetic acid = $\frac{givenmass}{molarmass}$ = $\frac{30}{60}$ = 0.5 mol
For complete neutralization,
Number of moles of $NaOH$ = number of moles of $C{H}_{3}COOH$
$1\times V=0.5$
Where, $V$ is voume of $NaOH$ required
Since, $V=0.5/1=1/x$
$\therefore x=2$