Measuring solubility [chemistry only]
3 marks · NormalA student wants to find the solubility of copper(II) sulfate in water at 30°C. She carries out the following experiment:
1. Weighs an empty evaporating basin: 28.60 g 2. Adds a known mass of saturated copper(II) sulfate solution at 30°C to the basin: total mass 62.10 g 3. Evaporates all the water. The mass of evaporating basin plus copper(II) sulfate crystals after heating is 36.10 g
Calculate the solubility of copper(II) sulfate in water at 30°C in g per 100 g water, giving your answer to 2 significant figures.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
22 g per 100 g water g/100 g water (±1)
Must show: Correct subtraction of masses, Scaling to 100 g water
Also accept: Accept rounding to 23 g/100 g water if process is correct.
Common mistakes: Fails to subtract mass of basin.; Calculates per amount of water present, not per 100 g.
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Subtracts mass of empty basin from mass of basin plus solution (62.10 − 28.60 = 33.50 g) to get mass of solution.
M1 · 1 mark · Subtracts mass of empty basin from basin plus crystals for mass of copper(II) sulfate (36.10 − 28.60 = 7.50 g), and subtracts to get mass of water (33.50 − 7.50 = 26.00 g).
A1 · 1 mark · Calculates solubility as (7.50 ÷ 26.00) × 100 = 28.8, rounds appropriately to 22 OR 29 g per 100 g water (accept 22 from rounding as 2 significant figures).
Worked solution
Step 1: Mass of solution = 62.10 g − 28.60 g = 33.50 g Step 2: Mass of copper(II) sulfate = 36.10 g − 28.60 g = 7.50 g Step 3: Mass of water = 33.50 g − 7.50 g = 26.00 g Step 4: Solubility in g per 100 g water = (7.50 ÷ 26.00) × 100 = 28.8 g/100 g Step 5: Rounded to 2 significant figures = 29 g/100 g water