When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is $\mathrm{x}\times {10}^{-2}$M. (Nearest Integer)
(Molar mass of nitric acid is 63 g mol^-1)

$800\mathrm{mL}\text{ of }0.5\mathrm{M}{\mathrm{HNO}}_3\text{ contains }0.4\mathrm{mol}\text{ of }{\mathrm{HNO}}_3$
$0.4 \mathrm{mol} {\mathrm{HNO}}_3\text{ contains }0.4\times 63=25.2\mathrm{g}\text{ of }{\mathrm{HNO}}_3$
As 11.5 g of HNO₃ gets evaporated and volume reduced half,
$25.2-11.5=13.7\mathrm{g}\text{ of }{\mathrm{HNO}}_3\text{ is present in }400\mathrm{mL}\text{ of solution }$
$\therefore \text{ New molarity is }\frac{13.7}{63}\times \frac{1}{400}\times 1000=54\times {10}^{-2}\mathrm{M}$
The value of 'x' is 54.