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)
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)
(Molar mass of nitric acid is 63 g mol–1)
Updated On: Oct 7, 2024
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Correct Answer: 54
Solution and Explanation
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 \(54\).
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