A sample of gold ore contains 22.0 g of gold metal per 100kg of ore. If gold has a market value of $350.00 per oz., what mass of ore, in tons, must be refined to get one million dollars worth of pure gold?
You want 1E6 dollars at $350/oz so you need 1E6/350 = ??ounces Au.
Convert ??ounces to grams (the problem doesn't specify Troy oz or Avoirdupois oz so you may need to ask). Multiply that by 100,000/22 because it isn't 100% gold in the ore,convert that to pounds then to tons. Post your work if you get stuck.
To find the mass of ore that needs to be refined to obtain one million dollars worth of pure gold, we need to follow these steps:
Step 1: Determine the market value of one gram of gold.
Since there are 28.35 grams in an ounce, we can convert the market value of gold from dollars per ounce to dollars per gram:
$350.00 per oz. ÷ 28.35 g per oz. = $12.34 per g
Step 2: Determine the mass of pure gold required to equal one million dollars.
One million dollars divided by the market value per gram of gold gives us:
$1,000,000 ÷ $12.34 per g = 81,046.92 g
Step 3: Convert the mass of pure gold to kilograms.
Since there are 1,000 grams in a kilogram:
81,046.92 g ÷ 1,000 g per kg = 81.05 kg
Step 4: Determine the mass of ore needed to obtain the desired mass of pure gold.
The mass of gold metal per 100 kg of ore is given as 22.0 g. Therefore, the mass of ore needed can be calculated as:
81.05 kg × (100 kg / 22.0 g) = 368.41 kg
Step 5: Convert the mass of ore to tons.
Since there are 1,000 kg in a ton:
368.41 kg ÷ 1,000 kg per ton = 0.37 tons
Therefore, approximately 0.37 tons of ore must be refined to obtain one million dollars worth of pure gold.