Each side of a metallic cube of mass 5.580 kg is measured to be 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as X × 10³ kg m³, where the value of X is: ____.

**Step 1 – Convert the side length to metres**

The cube side is given as 9.0 cm. Converting centimetres to metres (1 cm = 0.01 m) gives

$l=9.0\times 0.01=0.090 m$

Because the measurement “9.0 cm” contains two significant figures (the trailing zero after the decimal is significant), the length is kept to two sf.

**Step 2 – Find the volume of the cube**

For a cube,

$V=l^3$

Substituting the length:

$V={0.090}^3=7.29\times {10}^{-4} m{³}^{\mathrm{ }}$

Rounding the volume to the same number of significant figures as the length (two sf) gives

$V\approx 7.3\times {10}^{-4} m{³}^{\mathrm{ }}$

**Step 3 – Compute the density**

Density is mass divided by volume:

$\rho =m/V$

Using the given mass 5.580 kg and the volume from step 2:

$\rho =5.580/7.3\times {10}^{-4}=7.65\times {10}^3 \mathrm{kg}{m}^{\mathrm{-3}}$

The intermediate result 7.65 × 10³ kg m⁻³ contains three sf. The original length measurement limited the final answer to two sf, so we round the coefficient to two sf:

$\rho \approx 7.7\times {10}^3 \mathrm{kg}{m}^{\mathrm{-3}}$

**Result** – The density of the material, expressed as X × 10³ kg m⁻³, has X = **7.7**.