JEE MAIN - Chemistry (2022 - 26th July Morning Shift - No. 19)
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
54
Explanation
$$\mathrm{m}$$ moles of $$\mathrm{HNO}_{3}=800 \times 0.5$$
Moles of $$\mathrm{HNO}_{3}=400 \times 10^{-3} = 0.4 \text { moles }$$
Weight of $$\mathrm{HNO}_{3}=0.4 \times 63 \mathrm{~g} =25.2 \mathrm{~g} $$
Remaining acid $$=25.2-11.5 =13.7 \mathrm{~g} $$
$$ \begin{aligned} M &=\frac{13.7 \times 1000}{400 \times 63} \\ &=\frac{137}{252}=0.54 \\ &=54 \times 10^{-2} \end{aligned} $$
Moles of $$\mathrm{HNO}_{3}=400 \times 10^{-3} = 0.4 \text { moles }$$
Weight of $$\mathrm{HNO}_{3}=0.4 \times 63 \mathrm{~g} =25.2 \mathrm{~g} $$
Remaining acid $$=25.2-11.5 =13.7 \mathrm{~g} $$
$$ \begin{aligned} M &=\frac{13.7 \times 1000}{400 \times 63} \\ &=\frac{137}{252}=0.54 \\ &=54 \times 10^{-2} \end{aligned} $$
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