Question

Rhodium, a metal used as a catalyst and in jewelry, currently sells for $1500. per troy ounce. If 6.022 x 10^23 rhodium atoms have a mass of 102.9 grams, how many rhodium atoms could you buy for exactly one nickel

Answer 1

`2.19 * 10^(-4) \mathrm{Rh}`

Step-by-step explanation:

Atomic mass of Nickel = 58.69 g/mol

Mass of 1 mole of nickel atom = 58.69 gm

Now, mass of 1 nickel atom = Gram atomic mass/Avogadro number

=
`(58.69)/(6.023*10^(22)) =9.744*10^(23) g/atom`

Now, price of Rhodium =$1500 per troy ounce

Price of nickel in market = 0.577$/oz

No. of rhodium atoms needed to buy 1 nickel atom is

`9.744 * 10^(-23) \mathrm{g} * \frac{1 \mathrm{oz}}{31.1035 \mathrm{g}} * \frac{0.577 \mathrm{s}}{1 \mathrm{oz}} * \frac{1 \mathrm{oz}}{1500 \mathrm{s}}`

`\quad * \frac{31.1035 \mathrm{g}}{1 \mathrm{oz}} * \frac{6.023 * 10^(23) \mathrm{Rh} \text { atoms }}{102.9 \mathrm{g}}=2.19 * 10^(-4) \mathrm{Rh} \text { atoms }`