Redox Reactions And Electrochemistry Question 161
Question: 4 g of copper was dissolved in concentrated nitric acid. The copper nitrate solution on strong heating gave 5 g of its oxide. The equivalent weight of copper is [KCET 2004]
Options:
A) 23
B) 32
C) 12
D) 20
Show Answer
Answer:
Correct Answer: B
Solution:
In 5 gm CuO, 4 gm Cu and 1 gm O be present.
Element Wt. At Wt. $ Wt./At.Wt.\ne x $ Ratio
Cu 4 gm 63.5 4/63.5=.0625 $ \frac{.0625}{.0625}=1 $
O 1 gm 16 1/16 =.0625 $ \frac{.0625}{.0625}=1 $
Emperical formula = CuO of oxide In this oxide, oxidation no. of $ Cu=+2 $ Equivalent weight $ =\frac{\text{Molecular weight }}{\text{Oxidation no}\text{.}}=\frac{63.5}{2}\approx 31.75 $ but Equivalent weight should be an integeral no. = 32
BETA
