Logo
Overview

Practice: Density Problems

January 15, 2025
1 min read

Problem 1: The Golden Cube

A cube of gold has a side length of 2 cm. The density of gold is 19.3 g/cm319.3 \text{ g/cm}^3. What is the mass of the cube?

Solution:

  1. Find Volume: V=side×side×side=2×2×2=8 cm3V = side \times side \times side = 2 \times 2 \times 2 = 8 \text{ cm}^3.
  2. Use Formula: Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume}.
  3. Calculate: Mass=19.3×8=154.4 gMass = 19.3 \times 8 = 154.4 \text{ g}.

Problem 2: The Mystery Liquid

You have a beaker weighing 50g. You add 100 mL of a mystery liquid. The combined weight is 130g. What is the density of the liquid?

Solution:

  1. Find Liquid Mass: 130g (Total)50g (Beaker)=80g130\text{g (Total)} - 50\text{g (Beaker)} = 80\text{g}.
  2. Liquid Volume: 100 mL (which is 100 cm3100 \text{ cm}^3).
  3. Calculate: Density=80/100=0.8 g/cm3\text{Density} = 80 / 100 = 0.8 \text{ g/cm}^3.
  • Inference: This is likely alcohol or oil (less dense than water).

Problem 3: Relative Density

The relative density of Silver is 10.8. The density of water is 1000 kg/m31000 \text{ kg/m}^3. What is the density of Silver in SI units?

Solution:

  • Relative Density=Density of SubstanceDensity of Water\text{Relative Density} = \frac{\text{Density of Substance}}{\text{Density of Water}}.
  • 10.8=Density of Silver100010.8 = \frac{\text{Density of Silver}}{1000}.
  • Density of Silver=10.8×1000=10,800 kg/m3\text{Density of Silver} = 10.8 \times 1000 = 10,800 \text{ kg/m}^3.