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 $K_{2}C{r}_2{O}_7$ in given sample?

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

Answer 1

Answer: (A)

$K_{2}C{r}_2{O}_7+KI\to I_2+C{r}^{3+}$
$I_2+N{a}_2{O}_3\to I_2+S_4{O}_6^{2-}$
$N{a}_2{S}_2{O}_3+K_{2}C{r}_2{O}_7\to$
Meq of $N{a}_2{S}_2{O}_7\to$ Meq. of $K_{2}C{r}_2{O}_7$
$320\times N=15\times \frac{1}{20}N=\frac{1}{40}$
Meq. of $I_2=$ Meq. of Hypo
Meq. of $I_2=$ Meq. of KI
Meq of KI = Meq. of $K_{2}C{r}_2{O}_7$
$24\times =\frac{1}{40}=$Meq. of 25 mL $K_{2}C{r}_2{O}_7$
$Meq.of500mL{K}_{2}C{r}_2{O}_7=\frac{24}{40}\times \frac{500}{25}$
$\frac{w\times 6}{294}\times 1000=12w=0.588$
$\%$ purity $=\frac{0.588}{0.8}\times 100=73.5\%$