How many troy ounces of fine gold are in 75 grams of 18K gold?
⚡ Instant Calculated Answer
75 grams of 18K gold contains exactly 56.25 grams (1.8085 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $4,882.95 USD at $2,700/oz t.
1.8085 oz t
Primary Metric
56.25 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 18K gold: 18 / 24 = 0.7500 (or 750.0‰ millesimal fineness).
- Multiply gross weight by fineness: 75 g × 0.7500 = 56.25 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 56.25 / 31.1034768 = 1.8085 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 56.25 / 11.6638 = 4.8226 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 37.5 g | 28.125 g | 0.9042 oz t | $2,441.34 |
| 60 g | 45.0 g | 1.4468 oz t | $3,906.36 |
| 75 g | 56.25 g | 1.8085 oz t | $4,882.95 |
| 90 g | 67.5 g | 2.1702 oz t | $5,859.54 |
| 112.5 g | 84.375 g | 2.7127 oz t | $7,324.29 |
| 150 g | 112.5 g | 3.617 oz t | $9,765.90 |
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