$H_{3} C C H_{2} O H + 2 K_{2} C r_{2} O_{7} + 8 H_{2} S O_{4} \to$ $3 C H_{3} C O O H + 2 C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + 11 H_{2} O$

Equivalent mass of is $K_{2} C r_{2} O_{7}$ is__________.

\frac{Molar mass}{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{Molar mass}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{Molar mass}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{Molar mass}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{Molar mass}{6}$
The oxidation number of chromium changes from +6 to +3.
Hence, the change in the oxidation number per chromium atom is +3.
The change in the oxidation number of two chromium atoms is +6.
Thus, the equivalent weight of ${C r_{2} O_{7}}^{2 -} = \frac{Molar mass}{6}$ .
Hence, the correct option is $(A)$

Suggest Corrections

Similar questions

F e S_{2} + K M n O_{4} + H^{+} \to F e^{+ 3} + S O_{2} + M n^{+ 2} + H_{2} O

F e S_{2}

{3 C H}_{3} {C H}_{2} O H + 2 K_{2} {C r}_{2} O_{7} + 8 H_{2} S O_{4} ⟶ 3 {C H}_{3} C O O H + 2 {C r}_{2} {(S O_{4})}_{3}

+ {2 K}_{2} S O_{4} + 11 H_{2} O

{C H}_{3} {C H}_{2} O H

2 K_{2} C r_{2} O_{7} + 8 H_{2} S O_{4} \to 2 K_{2} S O_{4} + 2 C r_{2} (S O_{4})_{3} + 8 H_{2} O + 3 O_{2}

2 N a O H + H_{3} P O_{4} \to N a_{2} H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

\frac{M}{3}

= \frac{Molar mass}{n-factor}

{C r_{2} O_{7}}^{2 -} \to C r^{3 +}

{C r_{2} O_{7}}^{2 -} = \frac{Molar Mass}{3}

Join BYJU'S Learning Program