8g of sulphur is burnt to form SO₂ which is oxidised by chlorine water. The solution is treated with BaCl₂ solution. The number of moles of BaSO₄ precipitated is

Wt of Suplhur brunt in air to form SO₂ = 8g

$$\begin{gathered} \mathrm{S} + {\mathrm{O}}_2 \to {\mathrm{SO}}_2 \\ 32 \mathrm{gm}-1 \mathrm{mole} \\ 8 \mathrm{gm}-{\mathrm{x}}_1 \mathrm{moles} \\ {\mathrm{x}}_1=\frac{8}{32}\times 1 =0.25 \mathrm{mole} \end{gathered}$$

II:  liberated SO₂ is oxidized by Chlorine water

Cl₂ + H₂O → 2HCl + (O)

SO₂ + (O) → SO₃

SO₃ + H₂O → H₂SO₄

$$\begin{gathered} {\mathrm{Cl}}_2 +2{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2 \to {\mathrm{H}}_2{\mathrm{SO}}_4 +2 \mathrm{HCl} \\ 1 \mathrm{mole} \mathrm{of} {\mathrm{SO}}_2 -1 \mathrm{mole} {\mathrm{ofH}}_2{\mathrm{SO}}_4 \\ 0.25 \mathrm{moles} \mathrm{of} {\mathrm{SO}}_2 -{\mathrm{x}}_2 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4 \\ {\mathrm{X}}_2=\frac{0.25\times 1}{1} =0.25 \mathrm{moles} \end{gathered}$$

III:  H₂SO₄ solution is treated with BaCl₂

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4 +{\mathrm{BaCl}}_2\to {\mathrm{BaSO}}_4+2\mathrm{HCl} \\ 1 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4-1 \mathrm{mole} \mathrm{of} {\mathrm{BaSO}}_4 \\ 0.25 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4 -{\mathrm{x}}_3 \mathrm{moles} {\mathrm{ofBaSO}}_4 \end{gathered}$$

Number of moles of BaSO₄ precipitated = 0.25