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 $\times$ 10$-$2 M. (Nearest integer)
(Molar mass of nitric acid is 63 g mol$-$1)
Answer (integer)
54
Solution
$\mathrm{m}$ moles of $\mathrm{HNO}_{3}=800 \times 0.5$
<br/><br/>
Moles of $\mathrm{HNO}_{3}=400 \times 10^{-3} = 0.4 \text { moles }$
<br/><br/>
Weight of $\mathrm{HNO}_{3}=0.4 \times 63 \mathrm{~g}
=25.2 \mathrm{~g}$
<br/><br/>
Remaining acid $=25.2-11.5
=13.7 \mathrm{~g}$
<br/><br/>
$$
\begin{aligned}
M &=\frac{13.7 \times 1000}{400 \times 63} \\
&=\frac{137}{252}=0.54 \\
&=54 \times 10^{-2}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
This question is part of PrepWiser's free JEE Main question bank. 163 more solved questions on Some Basic Concepts of Chemistry are available — start with the harder ones if your accuracy is >70%.