A sample of ${NaClO}_{3}$ is converted by heat to $NaCl$ with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as $AgCl$ (molar mass of $AgCl : 143.5 g {mol}^{- 1}$ ) obtained will be

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0.54 g

0.35 g

0.48 g

0.41 g

answer is C.

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Detailed Solution

The reaction is ${NaClO}_{3} \underset{Δ}{⟶} NaCl + \frac{3}{2} O_{2}$

Amount of $O_{2}$ released $= \frac{0.16 g}{32 g {mol}^{- 1}} = 0.005 mol$

Mass of $AgCl$ precipitated $= (\frac{2}{3} \times 0.005 mol) (108 + 35.5) g {mol}^{- 1} = 0.48 g$

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