Verified Solution • 2025–26 Session moderate

QJEEM22Chdb3c: JEE Main 2022 Chemistry - Class 11 JEEM22Chdb3c Solution: When 800 mL of 0.5 M nitric acid is heated in a beaker, its…

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)

Step-by-Step Verified Solution

m\mathrm{m} moles of HNO3=800×0.5\mathrm{HNO}_{3}=800 \times 0.5

Moles of HNO3=400×10−3=0.4 moles \mathrm{HNO}_{3}=400 \times 10^{-3} = 0.4 \text { moles }

Weight of HNO3=0.4×63 g=25.2 g\mathrm{HNO}_{3}=0.4 \times 63 \mathrm{~g}=25.2 \mathrm{~g}

Remaining acid =25.2−11.5=13.7 g=25.2-11.5 =13.7 \mathrm{~g}

Mamp;=13.7×1000400×63amp;=137252=0.54amp;=54×10−2\begin{aligned}M &=\frac{13.7 \times 1000}{400 \times 63} \\&=\frac{137}{252}=0.54 \\&=54 \times 10^{-2}\end{aligned}

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