The density of gold is 19.32 g/cm^3. And ingot of gold measures 15 x 7.5 x 2.0 cm. How many gold atoms are in the ingot? Is the answer 1.33 x 10^25 atoms (with correct significant figures)?
2007-09-24 06:33:39 · 3 answers · asked by greddy_civic 1 in Science & Mathematics ➔ Chemistry
The volume of the ingot if V= L*W*H
V=15cm*7.5cm*2.0cm = 225 cm^3
density = 19.32 g/cm^3
V*density = grams gold ==> 225cm^3 * 19.32g/cm^3
= 4320 g ==> 21.93 moles
MW AU = 196.97 g/mole
21.93 moles (6.023*10^23 atoms/mole) = 1.32 * 10^25
Yep you are CORRECT!!
2007-09-24 06:49:41 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
You do the volume calc to get 225 cm^2
So there is about 4800 g of gold in ingot. Divide that by the g-atomic wt of gold (appx 200) and you have 24 g-atoms. So your answer looks reasonable EXCEPT that the ingot measurements are to two significant figures, so the answer should be also to two significant figures.
2007-09-24 07:35:51 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
Your ingot is 225 cm^3 so you multiply by the 19.32 to get 4347g then divide by 196.967amu to get 22.069686003 moles multiplied by 6.022x10^23 giving you 1.3290678759x10^25 which you then round to 1.33x10^25 with sig figs, so yes you are correct.
2007-09-24 06:45:15 · answer #3 · answered by Anonymous · 1⤊ 0⤋