From the graph, solubility of \(KClO _{3}\) at \(65^{\circ} C \Longrightarrow 6 moldm ^{-3}\),
that at \(25^{\circ} C =1\) moldm \(^{-3}\)
Mole present in solution at \(65^{\circ} C\)
\(=\left(\frac{6 \times 20}{100}\right) \Rightarrow 1.2\) mole
Mole present in solution at \(25^{\circ} C\)
\begin{aligned}
&=\left(\frac{1 \times 200}{1000}\right) \Rightarrow 0.2 \text { mole } \\
&\text { Precipitate }=1.2 mole -0.2 \text { mole } \\
&\Rightarrow \text { 1mole }
\end{aligned}