Question

(3) 10.5 (4) 0.5 In a given sample, oleum is labelled as 104.5%. The percentage of free Soz in it is (1) 4.5% (2) 10% (3) 20% (4) 40% 81. Weight of Na coz of 80% purity would be required

NEET/Medical Exams

Chemistry

Solution

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( mathrm{SO}_{3}+mathrm{H}_{2} mathrm{O} rightarrow mathrm{H}_{2} mathrm{SO}_{4} ) ( mathrm{x} / 80 mathrm{mole} ) of ( mathrm{SO}_{3} ) produces ( mathrm{x} / 80 mathrm{mole} ) of ( mathrm{H}_{2} mathrm{SO}_{4} ) weight of ( mathrm{H}_{2} mathrm{SO}_{4}=mathrm{x} / 80^{star} 98 ) Total weight of ( mathrm{H}_{2} mathrm{SO}_{4} ) present in solution after dilution ( x / 80^{star} 98+(100-x)=104.5 ) ( x=20 mathrm{g} ) Therefore, %age of free ( mathrm{SO}_{3} ) is ( 20 % ).