Question

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 $x × 10^{–2} M$. (Nearest integer)
(Molar mass of nitric acid is 63 g mol–1)

Answer 1

m moles of $HNO_3=800×0.5$
Moles of $HNO_3=400×10^{−3}=0.4$ moles
Weight of $HNO_3 = 0.4 × 63 g= 25.2$ g
Remaining acid $= 25.2 – 11.5= 13.7$ g
$M=\frac {13.7×1000}{400×63}$

$M=\frac {137}{252}$
$M=0.54$
$M=54×10^{−2}$
Given that, The molarity of the remaining nitric acid solution is $x×10^{−2} M$.
On comparing, $x= 54$

So, the answer is $54$.