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InterviewSolution
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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.A. `3.6%`B. `1.875%`C. `9%`D. `4.5%` |
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Answer» Correct Answer - B 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%` |
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