Question

In the standardisation of Na$_2S_2O_3$ using K$_2Cr_2O_7$ by iodometry, the equivalent weight of K$_2Cr_2O_7$ is

• A. (molecular weight)/2
• B. (molecular weight)/6
• C. (molecular weight)/3
• D. same as molecular weight

Answer 1

Answer: (B)

(molecular weight)/6

Answer 2

The following reaction occur between $S_2O_3^{2-} \, and \, Cr_2O_7^{2-}:$
$26H^+ +3S_2O_3^{2-}+4Cr_2O_7^{2-}\rightarrow 6SO_4^{2-}+8Cr^{3+}+13H_2O$
Change in oxidation number of $Cr_2O_7^{2-}$ per formula unit is 6 (it is
always fixed for $Cr_2O_7^{2-}$
Hence equivalent weight of $K_2Cr_2O_7 =\frac{Molecular \, weight}{6}$