Chemistry - Concentration Terms Question with Solution | TestHub

To refine the solution, we first calculate the molar masses:

Molar mass of SO₃ = $32+3\times 16=80$ g/mol

Molar mass of H₂O = $2\times 1+16=18$ g/mol

Molar mass of H₂SO₄ = $2\times 1+32+4\times 16=98$ g/mol

The reaction is: SO₃ + H₂O $\to$ H₂SO₄

109% oleum means 100 g of oleum yields 109 g of H₂SO₄ upon dilution.

Let 'x' be the mass of free SO₃ in 100 g of oleum.

Mass of H₂SO₄ initially present = $(100-x)$ g.

Mass of H₂SO₄ formed from 'x' g of SO₃ = $x\times \frac{98}{80}=1.225x$ g.

Total H₂SO₄ = $(100-x)+1.225x=109$ g

$100+0.225x=109$

$0.225x=9$

$x=\frac{9}{0.225}=40$ g

(P) Mass of free SO₃ = 40 g. (Matches with (1))

(Q) Mass of H₂SO₄ in 100 g oleum = $100-40=60$ g. (Matches with (2))

(R) Moles of H₂SO₄ = $\frac{60}{98}\approx 0.6122$ mol $\approx 0.61$ mol. (Matches with (3))

(S) Mass of water required to convert 100 g of 109% oleum to 100% H₂SO₄ is the water needed to react with 40 g SO₃.

Moles of SO₃ = $\frac{40}{80}=0.5$ mol.

From the reaction, 1 mole of SO₃ reacts with 1 mole of H₂O.

Moles of H₂O needed = 0.5 mol.

Mass of H₂O = $0.5\times 18=9$ g. (Matches with (4))

Therefore, the correct match is (P)-(1), (Q)-(2), (R)-(3), (S)-(4).