Question

Find the no. of copper atom present in 10g of 22 carat gold.​

Answer 1

Question

Find the no. of copper atom present in 10g of 22 carat gold.

Answer

first of all, find percentage of copper present in 22 carat gold.

percentage of copper in 22 carat gold = (24 - 22)/24 × 100 = 100/12 = 25/3 %

now,

weight of copper in 10g of 22 carat gold = 25/3 % of 10g = 25/3 × 10/100

= 10/12 g

now,

number of copper atoms in (10/12)g = number of mole of copper × Avogadro's number

= given weight/atomic mass × 6.023 × 10²³

= (10/12)/63.5 × 6.023 × 10²³

= (10 × 6.023)/(12 × 63.5) × 10²³

= 0.079 × 10²³

= 7.9 × 10²¹

Answer 2

ANSWER

There is 2 parts of, out of 24 parts, copper is present.

Mass of copper=

$= \frac{2}{24} \times 10g$

Number of moles of Cu

$= \frac{mass \: given}{molecular \: mass}$

$= \frac{2}{24} \times \frac{10}{63.5}$

No. of atoms of Cu= No. of moles×NA

$= \frac{2}{24} \times \frac{10}{63.5} \times 6.022 \times {10}^{23}$

$= 7.9 \times {10}^{21}$

THANKS