How much amount of `CuSO_(4)*5H_(2)O` is required for liberation of 2.54g of `I_(2)` when titrated with KI-

Options

A. 2.5g
B. 4.99g
C. 2.4g
D. 1.2g

Answer

B

Explanation

`2CuSO_(4)*5H_(2)O+4KI to Cu_(2)I_(2)+2K_(2)SO_(4) +I_(2)+10H_(2)O`
Molecular weight of `I_(2)=254g`
Molecular weight of `2CuSO_(4)*5H_(2)O`
`=[2(63.5+32+64)+10(18)]g=499g`
254g of `I_(2)` is liberated by 499g of `CuSO_(4)*5H_(2)O`
2.54g of `I_(2)` will be liberated by x g `CuSO_(4)*5H_(2)O`
`therefore x=(499)/(254)xx2.54=4.99g`g