1.

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

Answer»

`5.2` GM
`4.57` gm
`2.3 ` gm
None of the above

SOLUTION :`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`


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