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0.635 g Cu was dissolved in 5.0 mL hot 6...

0.635 g Cu was dissolved in 5.0 mL hot 60% `HNO_(3)` (sp. gr. = 1.5). When the reaction came to an end, the volume of the solution was adjusted to 250.0 mL, What is the normality of the solution with respect to `HNO_(3)` ? (Cu= 63.5)

A

0.256 N

B

0.126 N

C

0.324 N

D

0.425 N

Text Solution

Verified by Experts

The correct Answer is:
B
0.635 g Cu was dissolved in 5.0 mL hot 60% `HNO_(3)` (specific gravity = 1.5)
The following reaction occurs:
`Cu + 4HNO_(3) =Cu(NO_(3))_(2) + 2NO_(2) + 2H_(2)O`
Molecular weight of `HNO_(3) = (1 + 14 + 48) = 63`
In equivlent weight = 63/1 = 63.
63.5 g of Cu reacts with `4 xx 63 = 252 g` of `HNO_(3)`
`therefore 0.635` g of Cu reacts with `252/63.5 xx 0.635 g` of `HNO_(3) = 2.52 g` of `HNO_(3)`
`=1.98/63` g equilvaent of `HNO_(3)`
`therefore 1000 mL` of `HNO_(3)` contains `=1.98/63 xx 1000/250 = 0.126`g equivalent of `HNO_(3)`
The normality of the solution with respect to `HNO_(3)` is 0.126 (N)
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