2004 Section 1 Question 29

Concentration of initial 250ml (0.25l) = 30g/l
Moles = concentration x volume = 30 x 0.25 = 7.5 mol
Volume after spillage = 0.2 l
Moles after spillage = 30 x 0.2 = 6 mol

Volume of new NaCl sol = 0.05 l
Concentration of new NaCl sol = 20g/l
Moles of new NaCl solution = 20 x 0.05 = 1 mol
Total moles = 6+1 = 7 mol

Concentration = moles/volume = 7/0.25 = 28 g/l