A certain amount of copper has to be dissolved in nitric acid. Which of the following is required in lesser amount of $H N O_{3}$

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$20 % H N O_{3}$

$60 % H N O_{3}$

$90 % H N O_{3}$

Funning nitric acid

answer is B.

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Detailed Solution

The reaction of copper with dilute and concentrated nitric acids are as follows

$C u + 4 H N O_{3} \to C u {(N O_{3})}_{2} + 2 H_{2} O + 2 N O_{2}$

$3 C u + 8 H N O_{3} \to 3 C u {(N O_{3})}_{2} + 4 H_{2} O + 2 N O$

To dissolve one mole of copper metal, 4 moles of concentrated $H N O_{3}$ or fuming nitric acid or $90 % H N O_{3}$ is required. To dissolve 3 moles of copper metal 8 moles of dilute $H N O_{3}$ is required. This indicates that for the dissolution of one mole of copper the amount of $H N O_{3}$ required is $\frac{8}{3} = 2.67$