A fitting has mass 156.4 g. Water in a measuring cylinder rises from 48.0 cm³ to 67.5 cm³ when it is submerged. (a) Calculate density in g/cm³ and kg/m³. (b) Explain how to make the measurements reliably. (c) A trapped bubble occupies 1.2 cm³. Calculate corrected density and explain the original error direction. (d) Explain why dissolving material needs another method.

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Accepted Answer

8.02 g/cm³ = 8.02 × 10³ kg/m³; corrected density 8.55 g/cm³. A bubble makes apparent volume too large and density too small. The fitting volume is the water displacement, 67.5 − 48.0 = 19.5 cm³. It must be fully submerged without overflow because this difference is being equated to object volume. Density is ρ = m/V = 156.4/19.5 = 8.02 g/cm³. Since 1 g/cm³ equals 1000 kg/m³, the SI density is 8.02 × 10³ kg/m³. Zero the balance, dry the fitting before weighing, read the cylinder at eye level at the bottom of the meniscus and repeat. A narrower graduated cylinder can reduce the percentage volume uncertainty. The bubble is not metal, so corrected volume = 19.5 − 1.2 = 18.3 cm³. Corrected density = 156.4/18.3 = 8.55 g/cm³; the original oversized denominator biased density downward. If material dissolves, both specimen mass and liquid composition change, and final volume is not simply water plus original solid volume. Use an inert displacement liquid or geometric method instead.

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