(a)(i) - Neutralization of a base by an acid
- Direct combination of elements
(ii) Thermometer, water bath and stirrer
(b)(i) The solubility of K at 50\(^o\) equal to 0.58mol/dm\(^3\)
(ii) The solubility of L equals to 1.0mol/dm\(^{-3}\) at 57\(^o\)C
(iii) K is more soluble than L over the temperature range of 0\(^o\) to 31.5\(^o\)C
(iv) Molar mass of L = 101g
3.4gm of L per 250cm\(^3\) \(\to\) 3.4 \(\to\) 250X \(\to\) 1000X = \(\frac{1000}{250}\) x 3.4 = 13.6g/dm\(^3\)
At 20\(^o\)C the molar conc. of L = 0.34
The solubility 101 x 0.34 = 34.34g/dm\(^3\) at 20\(^o\)C. Since the concentration/solubility of 13.6g/dm\(^3\) is lower than the required solubility of 34.34g/dm\(^3\) at 20\(^o\)C, it is therefore not saturated ie. unsaturated