In a quality control analysis for sulphur impurity `0.56g` steel sample was burnt in a stream of oxygen and sulpphur was converted into `SO_(2)` gas. The `SO_(2)` was then oxidized to sulphate by using `H_(2)O_(2)` solution to which had been added `30mL` of `0.04M NaOH`. the equation for reaction is:
`SO_(2(g)) +H_(2)O_(2(aq)) +2OH_((aq))^(-) rarr SO_(4(aq))^(-2) +2H_(2)O_((1))`
`22.48 mL` of `0.024M HCI` was required to neutralize the base remaining after oxidation reaction. Calculate % of sulphur in given sample.

Options

A. `3.6%`
B. `1.875%`
C. `9%`
D. `4.5%`

Answer

B

Explanation

Meq of alkali added `=30 xx 0.04 = 1.2`
Meq. Of alkali left `=22.48 xx 0.024 = 0.54`
`:.` Meq. of alkali for `SO_(2)` and `H_(2)O_(2) = 1.2 - 0.54 = 0.66 =` M.eqts of alkali
1 moles of alkai `-=1` moles of `SO_(2)`
`:.` Weight of `S = (0.66)/(2) xx (32)/(1000) = 0.0105g`
`:. %` of `S = (0.0105)/(0.56) xx 100 = 1.875%`