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 ?

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75.3 %

73.5 %

none of these

36.75 %

answer is A.

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Detailed Solution

$K_{2} {Cr}_{2} O_{7} + KI \overset{}{\to} I_{2} + {Cr}^{3 +}$

$I_{2} + {Na}_{2} S_{2} O_{3} \overset{}{\to} I_{2} + S_{4} O_{6}^{2 -}$

${Na}_{2} S_{2} O_{3} + K_{2} {Cr}_{2} O_{7} \overset{}{\to}$

Meq of ${Na}_{2} S_{2} O_{3} \overset{}{\to} Meq . of K_{2} {Cr}_{2} O_{7}$

$30 \times N = 15 \times \frac{1}{20} N = \frac{1}{40} Meq . of I_{2} = Meq . of Hypo Meq . of I_{2} = Meq . of KI Meq of KI = Meq . of K_{2} {Cr}_{2} O_{7} 24 \times \frac{1}{40} = Meq . of 25 mL K_{2} {Cr}_{2} O_{7} Meq . of 500 mL K_{2} {Cr}_{2} O_{7} = \frac{24}{40} \times \frac{500}{25} \frac{w \times 6}{294} \times 1000 = 12 w = 0.588 % purity = \frac{0.588}{0.8} \times 100 = 73.5 %$