MCQ
The equivalent weight of ${K_2}C{r_2}{O_7}$ in acidic medium
- A$294$
- B$298$
- ✓$49$
- D$50$
✓
Answer
Correct option: C.
$49$
c
(c) ${K_2}C{r_2}{O_7} + 3{H_2}S{O_4}\, \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3\,(O) + 3{H_2}$
(c) ${K_2}C{r_2}{O_7} + 3{H_2}S{O_4}\, \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3\,(O) + 3{H_2}$
No. of electrons lossed $= 12 -6 = 6$
Equivalent weight $=\frac{M}{6} = \frac{{294}}{6} = 49$ .
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Which of the following notation of orbital is incorrect considering the sign of wave function?
Silica is heated with carbon in electric furnace to form :
The ionic radii of $\mathrm{K}^{+}, \mathrm{Na}^{+}, \mathrm{Al}^{3+}$ and $\mathrm{Mg}^{2+}$ are in the order:
→Which has the highest value of ionic radius ?
During mixed acid nitration of benzene Conc. $HNO_3$ behaves like
→If Avogadro number $N_A,$ is changed from $6.022 \times 10^{23}\ mol^{-1}$ to $6.022 \times 10^{20}\ mol^{-1},$ this would change
→$sp^3d^2$ hybrid orbitals have
→At ${20\,^o}C,$ the $A{g^ + }$ ion concentration in a saturated solution of $A{g_2}Cr{O_4}$ is $1.5 \times {10^{ - 4}}\,mole/litre.$ At ${20\,^o}C,$the solubility product of $A{g_2}Cr{O_4}$ would be
→$20\, ml$ of $0.2\, M \,HCN$ mix with $10\, ml$ of $0.2\, M\, NaOH$, then calculate $pH$ of resulting mixture. $pKa$ value of $HCN$ is $5$
→Product $B$ may be :
→