\(100.0 g\) of \(KClO _{3}\) was added to \(40.0 cm ^{3}\) of water to give a saturated solution at \(298 K\). If the solubility of the salt is \(20.0 moldm ^{-3}\) at \(298 K\), what percentage of the salt is left undisolved? \([ K =39, Cl =35.5, O =16]
A. \(80 \%\) B. \(60 \%\) C. \(5 \%\) D. \(2 \%\)
Correct Answer: D
Explanation
Solubility (moldm \(m ^{-3}\) ) \(=\frac{\text { mass of solute } \times 1000}{\left(\begin{array}{l}\text { mass of solvent } \times \text { molar } \\ \text { mass of solute }\end{array}\right)}\) \(=39+35.5+(3 \times 16)=122.5 g / mol\) Mass of salt dissolved (solute) \(=x\) Mass of solvent (water) \(=40 cm ^{3}\) \(=40 g\) \(20=\frac{x \times 100}{40 \times 122.5}, x=98 g\) \(\therefore\) at \(298 K\), out of \(100 g\) of \(KClO _{3}\) given, \(98 g\) will dissolve. Mass of undissolved \(KClO _{3}\) \(=100 g -98 g =2 g\) \(\%\) of salt undissolved \(=\frac{\text { mass of salt undissolved }}{\text { total mass of salt }} \times 100 \%\) \(=\frac{2 g }{100 g } \times 100 \%=2 \%\)