Question

5 g of oleum is diluted with water. It requires 533.16 ml of 0.1 M $Ba(OH{)}_2$ solution for complete neutralization. Find oleum strength (in %).

Answer 1

533.16 ml of 0.1 M $Ba(OH{)}_2=533.16\times 0.1\times 2meqv=meqvof{H}_{2}S{O}_4$
So, mmol of $H_{2}S{O}_4=533.16\times 0.1mmols$
So, mass of $H_{2}S{O}_4=\frac{533.16\times 0.1\times 98}{1000}gm$
So, oleum strength $=\frac{533.16\times 0.1\times 98}{1000}\times \frac{100}{5}=104.5$