# Question #7013e

Apr 22, 2017

Refer to the explanation

#### Explanation:

First, find the moles of $N a O H$ using the formula of $n = c \cdot v$
number of moles = $\frac{v o l u m e}{1000} \cdot c o n c e n t r a t i o n$

moles = $\frac{40}{1000} \cdot 0.2$

moles = 0.008

Then here is the balanced equation of the neutralisation reaction while having the unknown dibasic acid.

${H}_{2} X + 2 N a O H \to 2 {H}_{2} O + N {a}_{2} X$

Mole ratio is 1:2 so you divide the moles of the NaOH by 2.

$\frac{0.008}{2}$ =0.004

Then we use the formula moles = $\frac{m a s s}{{A}_{r} \mathmr{and} {M}_{r}}$

${M}_{r} = \frac{m a s s}{m}$

$\frac{0.2}{0.004}$ = 50

Relative molecular mass of acid A is 50