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Question

Equivalent weight of $$K_{2}Cr_{2}O_{7}$$ as oxidant in acidic medium is:


A
24.5
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B
49
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C
147
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D
296
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Solution

The correct option is B $$49$$
In $$K_2Cr_2O_7$$, oxidation state of $$Cr=+6$$
In acidic medium,
$$Cr^{+6}\longrightarrow Cr^{+3}$$
$$\therefore$$ Change of oxidation state of $$Cr=3$$
$$\therefore$$ Change of oxidation state of $$K_2Cr_2O_7=6$$
$$\therefore$$ Equivalent weight of $$K_2Cr_2O_7=\cfrac {Molecular\ weight}{6}$$
=$$\cfrac {294}{6}$$
=$$49$$

Chemistry

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