Weight of 50% pure Potassium chlorate required to liberate 33.6 litres of oxygen at STP is

$$\begin{gathered} 2{\mathrm{KClO}}_3\left(\mathrm{s}\right) \stackrel{∆}{\to } 2\mathrm{KCl}\left(\mathrm{s}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right); \\ 2\times 122.5 \mathrm{g} 3\times 22.4 \mathrm{L} \\ \text{'}\mathrm{X}\text{'} 33.6 \mathrm{L} \\ \mathrm{X}=122.5 \mathrm{g} ; \\ \mathrm{but} \mathrm{the} \mathrm{purity} \mathrm{of} {\mathrm{KClO}}_3 \mathrm{is} \mathrm{only} 50\% \mathrm{which} \mathrm{has} \mathrm{to} \mathrm{liberate} 33.6 \mathrm{L} \mathrm{of} {\mathrm{O}}_2 ; \\ \mathrm{Thus} \mathrm{weight} \mathrm{of} 50\% \mathrm{pure} {\mathrm{KClO}}_3 \mathrm{required} =\frac{100}{50}\times 122.5 \mathrm{g} = 245 \mathrm{g} \end{gathered}$$