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25.0 g of FeSO(4).7H(2)O was dissolved i...

25.0 g of `FeSO_(4).7H_(2)O` was dissolved in water containing dilute `H_(2)SO_(4)` and the volume was made up to 1.0 L. 25.0 " mL of " this solution requried 20 " mL of " an `(N)/(10)` `KMnO_(4)` solution for complete oxidation the percentage of `FeSO_(4)7H_(2)O` in the acid solution is

A

`78%`

B

`98%`

C

`89%`

D

`79%`

Text Solution

Verified by Experts

The correct Answer is:
C

`implies((W_(FeSO_(4).7H_(2)O)xx1000)/((278)/(1))=80`
`impliesW_(FeSO_(4).7H_(2)O)=22.24`
`implies% of FeSO_(4).6H_(2)O=(22.24)/(25)xx100%=88.96%`
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25 g of a sample of FeSO_(4) was dissolved in water containing dil. H_(2)SO_(4) and the volume made upto 1 litre. 25 mL of this solution required 20mL of N/ 10 KMnO_(4) for complete oxidation. Calculate % of FeSO_(4).7H_(2)O in given sample.

32g of a sample of FeSO_(4) .7H_(2)O were dissolved in dilute sulphuric aid and water and its volue was made up to 1litre. 25mL of this solution required 20mL of 0.02 M KMnO_(4) solution for complete oxidation. Calculate the mass% of FeSO_(4) .7 H_(2)O in the sample.

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25 grams of a sample of ferrous sulphate is dissolved in dilute sulphuric acd. By adding water, its volume is made up to 1 litre. 25 mL of this solution requries 20 mL of N//10 KMnO_(4) solution for oxidation. Calculate the percentage of FeSO_(4) . 7H_(2)O in the sample. Strategy: Percentage of FeSO_(4).7H_(2)O =("mass"_("ferrous sulphate"))/("mass of sample")xx100% To get the mass of ferrous suphate, we need to calculate equivalents of ferrous sulphate. By law of equivalrnce, the milli equivalents of KMnO_(4) must be equal to the miliequivalents of ferrous sulphate. We can find the milliequivalents of KMnO_(4) by taking the product of millilitres of solution and its normality.

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