Answer:
The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³
Explanation:
The percentage by mass of gold in the ore = 0.22%
The density of the ore = 8.0 g/cm³
The price of the gold = $818 per troy ounce
14.6 troy oz = 1.0 pound
1 lb = 454 g
Given that one troy ounce = $818
$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce
1 troy oz = 1.0/14.6 lb
100/818 troy oz = 100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb
1 lb = 454 g
250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g
$100 = 3.8015 g worth of gold
The mass, M, of the ore containing 3.8015 g of gold is given as follows;
0.22% of M = 3.8015 g
0.22/100 × M = 3.8015 g
M = 3.8015 g × 100/0.22 = 1727.933 g
The volume, V, of the ore containing 3.8015 g of gold is given as follows;
Density of ore = Mass of ore/(Volume of ore)
Volume of ore = Mass of ore /(Density of ore)
The density of the ore = 8.0 g/cm³
Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³
Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.
