MCQ
$4\,g$ 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
- A$23$
- ✓$32$
- C$12$
- D$20$
|
Element |
Wt. |
At Wt. |
$Wt./At.Wt. \ne x$ |
Ratio |
|
$Cu$ |
$4\,gm$ |
$63.5$ |
$4/63.5=0.0625$ |
$\frac{{.0625}}{{0.0625}} = 1$ |
|
$O$ |
$1\,gm$ |
$16$ |
$1/16 =0.0625$ |
$\frac{{.0625}}{{0.0625}} = 1$ |
Emperical formula $= CuO$ of oxide
In this oxide, oxidation no. of $Cu = + 2$
Equivalent weight $ = \frac{{{\rm{Molecular\, weight }}}}{{{\rm{Oxidation\, no}}{\rm{.}}}} = \frac{{63.5}}{2} \approx 31.75$ but Equivalent weight should be an
integeral no.$ = 32$
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$(A)$ $\mathrm{Sc}$ $(B) \mathrm{Cr}$ $(C) \mathrm{V}$ $(D)$ $\mathrm{Ti}$ $(E)$ $\mathrm{Mn}$
Choose the correct answer from the options given below: