14 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 themixture is 1.1 g cm–3. Assume that the mixing process does not change the total volume of theliquid.What is the volume of water added?A 40 cm3 B 44 cm3 C 52 cm3 D 80 cm3

Let's solve this step by step:

1. First, we know that the density (d) of a substance is its mass (m) divided by its volume (v). So, we can write this as d = m/v.

2. We know the volume of glycerine (40 cm3) and its density (1.3 g/cm3). So, we can find the mass of the glycerine by rearranging the formula to m = d*v. This gives us m = 1.3 g/cm3 * 40 cm3 = 52 g.

3. We also know that the density of the mixture is 1.1 g/cm3. The mass of the mixture is the mass of the glycerine plus the mass of the water. The volume of the mixture is the volume of the glycerine plus the volume of the water. So, we can write the density of the mixture as (mass of glycerine + mass of water) / (volume of glycerine + volume of water).

4. We can rearrange this to find the volume of water: volume of water = (mass of glycerine + mass of water) / density of mixture - volume of glycerine.

5. We know the mass of glycerine (52 g), the density of the mixture (1.1 g/cm3), and the volume of glycerine (40 cm3). We don't know the mass of the water, but we know that the density of water is 1.0 g/cm3, so the mass of the water is the same as the volume of the water.

6. Substituting these values into the formula gives us: volume of water = (52 g + volume of water) / 1.1 g/cm3 - 40 cm3.

7. Solving this equation for the volume of water gives us a volume of 52 cm3. So, the answer is C, 52 cm3.