A sample of a pure metal has a mass of 12.15 kilograms. The volume of the sample is 4,500 cubic...

1. TRANSLATE the problem information

• Given information:
  • Mass = 12.15 kg (already in correct units)
  • Volume = 4,500 cm³ (needs conversion to m³)
  • Formula: \(\mathrm{density = \frac{mass}{volume}}\)
  • Want answer in \(\mathrm{kg/m^3}\)
• What this tells us: We need to convert volume units before calculating density.

2. INFER the conversion approach

• Since we need cubic meters \(\mathrm{(m^3)}\), we must cube the linear conversion factor
• The problem gives us: \(\mathrm{1~meter = 100~centimeters}\)
• For volume: \(\mathrm{1~m^3 = (100~cm)^3 = 100 \times 100 \times 100~cm^3 = 1{,}000{,}000~cm^3}\)

3. SIMPLIFY the volume conversion

• Convert \(\mathrm{4{,}500~cm^3}\) to \(\mathrm{m^3}\):
• Volume in \(\mathrm{m^3}\):

\(\mathrm{Volume~in~m^3 = 4{,}500~cm^3 \div 1{,}000{,}000~cm^3/m^3}\)

\(\mathrm{= 0.0045~m^3}\)

4. SIMPLIFY the density calculation

• Apply density formula: \(\mathrm{density = \frac{mass}{volume}}\)
• Density calculation:

\(\mathrm{Density = 12.15~kg \div 0.0045~m^3}\)

\(\mathrm{Density = 2{,}700~kg/m^3}\) (use calculator)

Answer: D) 2,700

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students use the linear conversion factor (100) instead of the cubic conversion factor \(\mathrm{(100^3 = 1{,}000{,}000)}\) for volume conversion.

They might calculate:

\(\mathrm{4{,}500~cm^3 \div 100 = 45~m^3}\)

\(\mathrm{density = 12.15 \div 45 = 0.27~kg/m^3}\)

This may lead them to select Choice A (0.27).

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the need for cubic conversion but make arithmetic errors in the final calculation or unit conversion.

For example, they might get the volume conversion right \(\mathrm{(0.0045~m^3)}\) but then calculate \(\mathrm{12.15 \div 0.0045}\) incorrectly, possibly getting 270 instead of 2,700.

This may lead them to select Choice C (27) after dropping a decimal place.

The Bottom Line:

This problem tests whether students understand that volume conversions require cubing the linear conversion factor. The key insight is recognizing that \(\mathrm{cm^3}\) to \(\mathrm{m^3}\) conversion uses \(\mathrm{100^3}\), not just 100.