Saved Bookmarks
| 1. |
25.0 cm^(3) of a solution containing 15.0 g of a partially oxidisied sample of green vitrion (FeSO_(4).7H_(2)O) per litre required 20.0 cm^(3) mL of 0.01 M potassium dichromate solution for oxidiation in aidic medium find out the percentage purity of the given sample of green vitriol |
|
Answer» Solution :Step 1 To write balaced EQUATION for the redox reaction `K_(2)Cr_(2)+4 H_(2)SO_(4) rarr K_(2)SO_(4)+Cr_(2)(sO_(4))_(3)+4H_(2)O+3O` `2FeSO_(4)+H_(2)SO_(4)+OrarrFe_(2)(SO_(4))+H_(2)O[xx3` `K_(2)Cr_(2)O_(7)+6FeSO_(4)+7H_(2)SO_(4)rarr K_(2)SO_(4)+Cr_(2)SO_(4)^(3)+3Fe_(2)SO_(4)^(3)+7H_(2)O` From the above equation 1 MOLE of `K_(2)Cr_(2)O_(7)=6` MOLES of `FeSO_(4)` Step 2 To find percentage purity of green vitroit Let `M_(1)` be the molarity of the oxidised sample of green vitril applying molarity eqaution we have `(M_(1)xx25)/(6)(FeSO_(4))=(20xx0.1)/(1)(K_(2)Cr_(2)O_(7))` `M_(1)=(20xx0.01xx6)/(25)=0.048 M` Mol wt of `FeSO_(4)7H_(2)O=56+32+4xx16+7xx18=278` Wt of pure `FeSO_(4), 7H_(2)O=278xx0.048=13.344 g L^(-1)` % purity of green vitriol `=(13.444)/(15)xx100=88.96` |
|