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\(gdm^{-3}\)
The correct answer is: B
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\(gdm^{-3}\)