1. Initial moles of nitric acid

Given:

• Initial volume $= 800\ \text{mL} = 0.8\ \text{L}$
• Initial molarity $= 0.5\ \text{M}$

So initial moles of $\text{HNO}_3$ are

$n_i = M \times V = 0.5 \times 0.8 = 0.4\ \text{mol}$

2. Moles of nitric acid evaporated

Mass of nitric acid evaporated $= 11.5\ \text{g}$

Molar mass of $\text{HNO}_3 = 63\ \text{g mol}^{-1}$

$n_{\text{evap}} = \frac{11.5}{63} \approx 0.18254\ \text{mol}$

3. Moles of nitric acid remaining

$n_f = 0.4 - 0.18254 = 0.21746\ \text{mol}$

4. Final volume of solution

The volume is reduced to half:

$V_f = \frac{800}{2}\ \text{mL} = 400\ \text{mL} = 0.4\ \text{L}$

5. Final molarity

$M_f = \frac{n_f}{V_f} = \frac{0.21746}{0.4} = 0.54365\ \text{M}$

This is approximately

$0.54\ \text{M} = 54 \times 10^{-2}\ \text{M}$

So,

$x = 54$

6. Comparison with stored answer

Stored correct answer $= 54$

This matches our derived answer.