The density of water is 1.0 g cm-3 and the density of glycerine is 1.3 g cm-3.

Question 21

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

Reference: Past Exam Paper – June 2018 Paper 12 Q10

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 = m_w

Volume of water = V_w

Density of glycerine = 1.3 g cm^-3

Mass of glycerine = m_g

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, m_g = 1.3 × 40 = 52 g

Density of mixture = total mass / total volume

Density of mixture = 1.1 g cm^-3

Total mass of mixture = m_w + m_g = m_w + 52

The mixing process does not change the total volume of the liquid.

Total volume of mixture = V_w + 40

Density = total mass / total volume

1.1 = (m_w + 52) / (V_w + 40)

V_w + 40 = (m_w + 52) / 1.1       ------------------- (1)

Consider the density of water.

Density = mass / volume

1.0 = m_w / V_w

m_w = 1.0 V_w = V_w                   ------------------- (2)

Replace m_w by V_w in equation (1).

V_w + 40 = (V_w + 52) / 1.1

V_w + 40 = V_w/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