# Question #b4f98

Feb 3, 2016

${M}_{r} = 55.25$

#### Explanation:

I'll refer to the acid as $H A$.

$H A + N a O H \rightarrow N a A + {H}_{2} O$

This tells us that 1 mole of $N a O H$ will neutralise 1 mole $H A$.

We can find the number of moles of $N a O H$ if we can find its volume.

Density $\rho$ is given by:

$\rho = \frac{m}{v}$

$\therefore v = \frac{m}{\rho} = \frac{32.68}{0.9705} = 33.673 \text{ml}$

Since $c = \frac{n}{v}$ then we can find the number of moles of $N a O H \Rightarrow$

${n}_{N a O H} = c \times v = 1 \times \frac{33.673}{1000} = 3.3673 \times {10}^{- 2}$

Because the acid and alkali react in a 1:1 ratio we can say that the number of moles of $H A$ must be the same.

$\therefore {n}_{H A} = 3.3673 \times {10}^{- 2}$

Now we need to find the mass of $H A$ used.

I will use the %w/v value of 62.02 which @Stefan found in another version of the question.

$\therefore {m}_{H A} / {V}_{s o \ln} = 0.6202$

You stated that 3ml of solution was used so:

${m}_{H A} = 0.6202 \times 3 = 1.8606 \text{g}$

$\therefore 3.3673 \times {10}^{- 2} \text{mol}$ weighs $1.8606 \text{g}$

So 1 mole weighs:

$\frac{1.8606}{3.3673 \times {10}^{- 2}} = 55.25 \text{g}$

${M}_{r} = 55.25$