Q#21 (Past Exam Paper – June 2018 Paper 12 Q10)
The density of water is 1.0 g cm-3 and the density of glycerine is 1.3 g cm-3.
Water is added to a measuring cylinder containing 40 cm3 of glycerine so that the density of the mixture is 1.1 g cm-3. Assume that the mixing process does not change the total volume of the liquid.
What is the volume of water added?
A 40 cm3 B 44 cm3 C 52 cm3 D 80 cm3
Solution:
Answer: D.
When the water and the glycerine are added, the density of the mixture is obtained by
Density = total mass / total volume
The following data are known.
Density of water = 1.0 g cm-3
Mass of water = mw
Volume of water = Vw
Density of glycerine = 1.3 g cm-3
Mass of glycerine = mg
Volume of glycerine = 40 cm3
The mass of the glycerine can be obtained from these two values.
Density = mass / volume
Mass = Density × Volume
Mass of glycerine, mg = 1.3 × 40 = 52 g
Density of mixture = total mass / total volume
Density of mixture = 1.1 g cm-3
Total mass of mixture = mw + mg = mw + 52
The mixing process does not change the total volume of the liquid.
Total volume of mixture = Vw + 40
Density = total mass / total volume
1.1 = (mw + 52) / (Vw + 40)
Vw + 40 = (mw + 52) / 1.1 ------------------- (1)
Consider the density of water.
Density = mass / volume
1.0 = mw / Vw
mw = 1.0 Vw = Vw ------------------- (2)
Replace mw by Vw in equation (1).
Vw + 40 = (Vw + 52) / 1.1
Vw + 40 = Vw/1.1 + 52/1.1
Vw – Vw/1.1 = 47.27 - 40
Vw (1 – 1/1.1) = 7.27
Vw = 7.27 / (1 – 1/1.1) = 79.97 = 80 cm3
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