Question

The equivalent weight of $K_{2}Cr_{2}O_{7}$ in acidic medium.

• A. 294
• B. 298
• C. 49
• D. 50

Answer 1

Answer: (C)

49

Answer 2

$K_{2}Cr_{2}O_{7} + 3H_{2}SO_{4} \rightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + 3(O) + 3H_{2}$

No. of electrons lossed = 12 – 6 = 6

∴ Equivalent weight = $\frac{M}{6} = \frac{294}{6} = 49.$