WAEC - Chemistry (2017 - No. 34)

If 0.2g of a salt is required to saturate 200\(cm^{3}\) of water at room temperature, what is the solubility of the salt?
0.2\(gdm^{-3}\)
1.0\(gdm^{-3}\)
2.0\(gdm^{-3}\)
5.0\(gdm^{-3}\)

Explanation

Solubility in \(gdm^{-3}\) = \(\frac{mass of salt in gram}{volume of liquid in dm^{3}}\)

200\(cm^{3}\) = \(\frac{200}{1000} dm^{3}  = 0.2dm^{3}\)

Solubility = \(\frac{0.2g}{0.2dm^{3}}\)

= 1.0\(gdm^{-3}\)

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