Answer:
the correct answer is 36
Step-by-step explanation:
The two substances are sodium chloride and lead nitrate
In case of sodium chloride;
mass of water = 100 kg
solubility of sodium chloride = 54 g / 100 g of water
Therefore the mass of sodium chloride required for full saturation,
L = (54/100) x 100 kg = 54 kg
In case of lead nitrate;
mass of water = 200 kg
mass of lead nitrate = 36 kg
if the solution is fully saturated, solubility of lead nitrate (g per 100 g of water)
k = (36/200) x 100 g = 18 g
k can be also written as 18 kg per 100 kg of water.
thus the difference in k and l = (54-18) = 36.