Question

(a) 109% oleum means 100 g oleum (H₂SO₄ + SO₃) exactly requires 9 g water to produce exactly 109 g of pure H₂SO₄.

$\underset{18\text{ g}}{H_{2}O} + \underset{\text{80 g}}{SO_{3}} \rightarrow H_{2}SO_{4}$

∴ 9 g → 40 g

∴ Percentage of free SO₃ = 40%

(b) 1 g of oleum contains 0.4 g of SO₃ and hence, 0.6 g of H₂SO₄.

$\text{Now, }n_{eq}SO_{3} + n_{eq}H_{2}SO_{4} = n_{eq}NaOH $$${\text{Or }\frac{0.4}{80} \times 2 + \frac{0.6}{98} \times 2 = \frac{V \times 0.5}{1000} \times 1 \Rightarrow V = 44.49\text{ ml} }{(c)n_{eq}SO_{3} + n_{eq}H_{2}SO_{4} = n_{eq}Ba(OH)_{2} }{\text{Or }\frac{2.0}{80} \times 2 + \frac{3.0}{98} \times 2 = \frac{V \times 0.1}{1000} \Rightarrow V = 111.22\text{ ml}}$$

(d) 100 g of oleum requires 9g of water to give 109 g of H₂SO₄. Hence, the final solution contains 109 g of H₂SO₄ and 491 g of water.

$\therefore\text{ Molality = }\frac{109/98}{491} \times 1000 = 2.265\text{ m}$