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0.80 g of sample of impure potassium dic...

`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

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(2)Cr_(2)O_(7)+KI rarr I_(2)+Cr^(3+)`
`I_(2)+Na_(2)S_(2)O_(3) rarr I_(2)+S_(4)O_(6)^(2-)`
`Na_(2)S_(2)O_(3)+K_(2)Cr_(2)O_(7) rarr`
Meq. Of `Na_(2)S_(2)O_(3) rar` Meq. Of `K_(2)Cr_(2)O_(7)`
`30xxN=15xx(1)/(20)N=(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)`
`24xx(1)/(40)=` Meq. of `25mL K_(2)Cr_(2)O_(7)`
Meq. of `500 mL K_(2)Cr_(2)O_(7)=(24)/(40)xx(500)/(25)`
`(wxx6)/(294)xx1000=12`, `w=0.588`
`% purity=(0.588)/(0.8)xx100=73.5%`
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Knowledge Check

  • 25 ml of H_(2)O_(2) solution was added to the excess of acidified Kl solution . The iodine so liberated required 40 ml of 0.1 N sodium thiosulphate solution. Calculate the normality of H_(2)O_(2) solution :

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