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Question

In a quality control analysis for sulphur impurity, 5.6 g of steel sample was burnt in a stream of oxygen. Sulphur was converted into SO2 gas. The SO2 was then oxidized to sulphate by using H2O2 solution to which had been added 30 mL of 0.04 M NaOH. The equation for reaction is as follows:


SO2(g)+H2O2(aq)+2OH(aq)SO24(aq)+2H2O(l)

22.48 mL of 0.024 M HCl was required to neutralize the base remaining after oxidation reaction. Calculate the percentage of sulphur in the given sample.

A
0.16%
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B
0.14%
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C
0.22%
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D
None of the above
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Solution

The correct option is B None of the above
Meq. of alkali added =30×0.04=1.2
Meq. of alkali left =22.48×0.024=0.54
Meq. of alkali used for SO2 and H2O2=1.20.54=0.66

Mass of alkali used =0.66×401000=0.0264 g

80 g NaOH reacts with 64 g of SO2.

0.0264 g NaOH reacts with 64×0.026480=0.021 g

Now, 64 g SO2 required =32 g S

0.021 g SO2 required =32×0.02164=0.0105 g

Percentage of S=0.01055.6×100=0.1875%

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