How many troy ounces of fine gold are in 10 grams of 24K gold?
⚡ Instant Calculated Answer
10 grams of 24K gold contains exactly 9.999 grams (0.3215 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $868.05 USD at $2,700/oz t.
0.3215 oz t
Primary Metric
9.999 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 24K gold: 24 / 24 = 0.9999 (or 999.9‰ millesimal fineness).
- Multiply gross weight by fineness: 10 g × 0.9999 = 9.999 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 9.999 / 31.1034768 = 0.3215 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 9.999 / 11.6638 = 0.8573 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 5 g | 5.0 g | 0.1608 oz t | $434.16 |
| 8 g | 7.999 g | 0.2572 oz t | $694.44 |
| 10 g | 9.999 g | 0.3215 oz t | $868.05 |
| 12 g | 11.999 g | 0.3858 oz t | $1,041.66 |
| 15 g | 14.998 g | 0.4822 oz t | $1,301.94 |
| 20 g | 19.998 g | 0.643 oz t | $1,736.10 |
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