In the standardization of N a 2 S 2 O 3 using K 2 Cr 2 O 7 by iodometry- the equivalent weight of K 2 Cr 2 O 7 is-Molecular weightA-B- Molecular weight - Molecular weightC- Molecular -3D- Same as its molecular weight

The correct option is B Molecular weight6The reaction between nbsp;Na_2 S_2 O_3 and K_2 Cr_2 O_7 occurs as follow: S_2 O_3 ^2 +Cr_2 O_7 ^2 → SO_4^2 +Cr^3+ Th ...

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Standard XII

Chemistry

Potassium Dichromate

In the standa...

Question

In the standardization of $N a_{2} S_{2} O_{3}$ using $K_{2} C r_{2} O_{7}$ by iodometry, the equivalent weight of $K_{2} C r_{2} O_{7}$ is:

\frac{Molecular weight}{2}

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\frac{Molecular weight}{6}

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\frac{Molecular weight}{3}

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Same as its molecular weight

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Solution

The correct option is B $\frac{Molecular weight}{6}$
The reaction between $N a_{2} S_{2} O_{3}$ and $K_{2} C r_{2} O_{7}$ occurs as follow:
$S_{2} O_{3}^{2 -} + C r_{2} O_{7}^{2 -} \to S O_{4}^{2 -} + C r^{3 +}$
The oxidation state of chromium changes from $+ 6$ to $+ 3$ . The net change in the oxidation number per formula unit is 6 or n-factor is 6.
$∴ Equivalent weight = \frac{Molecular weight}{6}$

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N a_{2} S_{2} O_{3}

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In the standardization of Na2S2O3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is:

In the standardization of $N a_{2} S_{2} O_{3}$ using $K_{2} C r_{2} O_{7}$ by iodometry, the equivalent weight of $K_{2} C r_{2} O_{7}$ is: