Chemistry - Complete P-block Question with Solution | TestHub

Let mass of oleum sample =$100g$.

∴Mass of SO₃ = $40g$

∴Mass of S₂SO₄ = $60g$

Mass of H₂O required for free SO₃ in $100g$ oleum =$\frac{40}{80}\times 18=9g$

Percentage labelling = $(100+9)\%=109\%$ Ans. (A)

For $200g$ oleum sample, $18g$ H₂O is required, but $9g$ water is added.

For $100g$, $4.5g$ water is added.

Mass of new oleum sample = $104.5g$.

$104.5g$ sample requires more H₂O = $4.5g$.

∴$100g$ requires more H₂O= $\frac{4.5}{104.5}\times 100$

∴ % labelling of new oleum= $\left(100+\frac{4.5\times 100}{104.5}\right)\%=104.3\%$