Question

if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius

Answer 1

The mass of ore required is 21 700 t.

r = 750 cm

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3} \pi (750 cm)^{3}$ = 1.767 × 10⁹ cm³

The density of lead is 11.34 g/cm³.

So mass of lead sphere = 1.767 × 10⁹ cm³ × $\frac{11.34 g}{1 cm^{3} }$ = 2.004 ×10¹⁰ g

2.004 ×10¹⁰ g × $\frac{1 kg}{1000 g}$ = 2.004 × 10⁷ kg

2.004 × 10⁷ kg × $\frac{1 t}{1000 kg}$ = 2.004 × 10⁴ t

92.5% efficiency means 92.5 t Pb per 100 t of ore.

Mass of ore = 2.004 × 10⁴ t Pb ×$\frac{100 tore}{92.5 t Pb}$ = 2.17 × 10⁴ t ore = 21 700 t ore