MCQ
The equivalent weight of ${K_2}C{r_2}{O_7}$ in acidic medium
- A$294$
- B$298$
- ✓$49$
- D$50$
No. of electrons lossed $= 12 -6 = 6$
Equivalent weight $=\frac{M}{6} = \frac{{294}}{6} = 49$ .
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$\begin{array}{*{20}{c}}
{C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2} - COOH}
\end{array}\xrightarrow{\Delta }(A)\xrightarrow{{C{H_3} - C{H_2}N{H_2}/\Delta }}(B)$
$(B)$ will be :
