how many atoms of gold are present in 0.6 gm of 18 caral gold? ( 24 caral gold in 100% pure) atomic weight of gold = 197?​

We are given that the ring is made up of 20 carrat gold. 20 carrat gold means 20 parts of gold and 4 parts of other metal (usually copper or silver). Thus, the percentage of gold in 20 carrat gold will be

= (20 / 24) X 100

= 83.333 %

We can now calculate the mass of gold in the ring. We are given that the ring weighs 300 mg. So, the mass of gold in ring will be

= (83.333 / 100) X 300

= 250 mg

Thus, the mass of gold in 300 mg of the ring is 250 mg.

Now the molar mass of gold is 197 g. This means that there are 6.023 X 1023 atoms of gold in 197g of gold. Thus

197g of gold = 6.023 X 1023 atoms of gold

1g of gold = (6.023 X 1023 ) / 197 atoms of gold

250 mg or 0.25g of gold = (6.023 X 1023 X 0.25) / 197 atoms of gold

= 7.643 X 1020 atoms of gold

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2