`NalO_(3)` reacts with `NaHSO_(3)` according to equation
` lO_(3)^(-) + 3HSO_(3)^(-) to l^(-) + 3H^(+) + 3SO_(4)^(2-)`
The weight of `NaHSO_(3)` required to react with 100 ml of solution containing `0.58` gm of `NalO_(3)` is

Options

A. `5.2` gm
B. `4.57` gm
C. `2.3 ` gm
D. None of the above

Answer

B

Explanation

`W_(NaHSO_(3))/M_(NaHSO_(3)) xx 2 xx 1000 = (0.58)/198 xx 6 xx 100`
`W_(NaHSO_(3)) = (0.58 xx 6 xx 1000 xx 104)/(198 xx 1000) = 4.57`