1.

5.6 g of a steel sample containing sulphur impurity was burnt in oxygen. SO_2 , so produced, was then oxidised to sulphate by H_2O_2solution to which 30 mL of 0.004 M NaOH solution had been added. 22.48 mL of 0.024 M HCl was required to neutralize the base remaining after oxidation reaction. Calculate percentage of S in the given sample of steel.

Answer»

Solution :`SO_2 + H_2O_2 + 2OH^(-) = SO_4^(2-) + 2H_2O`
m mol of S = MMOL of `SO_2 = 1/2 XX ` m mol of `OH^(-)= (30 xx 0.04) - 22.48 xx 0.024)`


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