Question

$0.80$ g of sample of impure potassium dichromate was dissolved in water and made up to $500$ mL solution. $25$ mL of this solution treated with excess of $KI$ in acidic medium and $I_{2}$ liberated required 24 mL of a sodium thiosulphate solution. 30 mL of this sodium thiosulphate solution required 15 mL of $N/20$ solution of pure potassium dichromate. What was the percentage of $K_{2} Cr_{2} O_{7}$ in given sample?

• A. $73.5\%$
• B. $75.3\%$
• C. $36.75\%$
• D. none of these

Answer 1

Answer: (A)

$K_{2}Cr_{2}O_{7} + KI \to I_{2}+ Cr^{3+}$
$I_{2} + Na_{2} O_{3} \to I_{2} + S_{4} O^{2-}_{6}$
$Na_{2}S_{2}O_{3} + K_{2}Cr_{2}O_{7} \to$
Meq of $Na_{2}S_{2}O_{7} \to $ Meq. of $K_{2} Cr_{2} O_{7}$
$320 \times N =15 \times \frac{1}{20} N =\frac{1}{40}$
Meq. of $I_{2} =$ Meq. of Hypo
Meq. of $I_{2} =$ Meq. of KI
Meq of KI = Meq. of $K_{2} Cr_{2} O_{7}$
$24\times=\frac{1}{40} =$Meq. of 25 mL $K_{2} Cr_{2}O_{7}$
$Meq. of 500 mL K_{2}Cr_{2}O_{7}=\frac{24}{40} \times\frac{500}{25}$
$\frac{w \times6}{294} \times1000 =12 w=0.588$
$\%$ purity $=\frac{0.588}{0.8} \times100 =73.5\%$