MCQ
The normality of $10\,lit.$ volume hydrogen peroxide is
- A$0.176$
- B$3.52$
- ✓$1.78$
- D$0.88$
$1 M$ ${H_2}{O_2}$solution $ = 2N = 34\,gm/litre = 11.2$
So Normality $ = \frac{{2 \times 10}}{{11.2}} = 1.78$
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$NH_2-CH_2-CH_2-NH_2 +$ $\begin{array}{*{20}{c}}
{COO{C_2}{H_5}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{COO{C_2}{H_5}}
\end{array}$ $\xrightarrow{{Pyridne}}$ $[X]$
$[X]$ will be :
[Given : $\sqrt{3}=1.73$ ]