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Grade Select GradePhysical Chemistry

what will be the volume of o2 at ntp liberated by 5A current flowing for 193 sec through acidulated water.

Profile image of Mrinal Sharma
11 Years agoGrade Select Grade
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1 Answer

Profile image of Pankaj
11 Years ago
Mass of Oxygen liberated(w) = EQ/(96500) = 8x965/965000
32 g of O2 = 22400mL
8x965/965000 g of O2 = [22400/32 ][ 8x965/965000] = 56mL