Question

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

Answer 1

Answer: 54

$n_{HN{O}_3}=0.5\times 0.8$
$=0.4$ mole
${\left(n_{HN{O}_3}\right)}_{\text{remains }}=0.4-\frac{11.5}{63}$
$=0.4-0.1825$
$=0.2175$
Molarity $=\frac{0.2175}{400}\times 1000$
$=\frac{0.2175}{0.4}$
$=0.5437mole/lit.$
$\simeq 0.54mole/lit.$
$=54\times 10^{-2}mol/lit.$