
The number of gold atoms in $300mg$ of a gold ring of $20 - $ carat gold. $\left( {Au = 197} \right)$ are:
A.$4.5 \times {10^{20}}$
B.$6.8 \times {10^{15}}$
C.$7.6 \times {10^{20}}$
D.$9.5 \times {10^{20}}$
Answer
566.4k+ views
Hint:Gold exists in various forms starting from the lowest as $10K$ and as high as $24K$ . We can calculate the number of atoms of gold using Avogadro’s number. In order to find the number of atoms we will calculate the weight of gold present in $20 - $ carat gold of a gold ring.
Complete step by step answer:
Pure gold is $24K$ or $24$ carat.
If gold is $20$ carat this means that it contains $20$ carat of pure gold and $4$ carat is of some other metal.
So if $300mg$ of gold ring is $20 - $ carat gold.
It means that $300mg$ contains $20$ carat of pure gold and $4$ carat is of some other metal.
Substituting the values we get,
Therefore weight of pure gold in $300mg$ gold ring $ = 300 \times \dfrac{{20}}{{24}}$
Weight of pure gold in $300mg$ gold ring $ = 250mg$
Weight of pure gold in $300mg$ gold ring $ = 0.25g$
Given data :
Molecular weight of pure gold $ = 197$
To find number of moles of gold we will use the formula: $M = \dfrac{W}{{MW}}$ ,where, $M = $ Number of moles, $W = $ Weight of a substance, $MW = $ Molecular weight.
Soln:
$M = \dfrac{W}{{MW}}$
Substituting the weight of gold and molecular weight of gold we will get,
$M = \dfrac{{0.25}}{{197}}$
$M = 0.00127$
Therefore, the number of moles present in $0.25g$ of gold is $0.00127$ .
Now we will calculate number of atoms of gold present
We will use the formula given below:
Number of atoms $ = n \times {N_A}$
Where, $n = $ number of moles, ${N_A} = $ Avogadro’s number
Given data:
Number of moles of gold $ = 0.00127$
Avogadro’s number ${N_A} = 6.023 \times {10^{23}}$
Number of atoms of gold $ = n \times {N_A}$
Substituting the values in the above formula we get,
Number of atoms of gold $ = 0.00127 \times 6.023 \times {10^{23}}$
Number of atoms of gold $ = 0.00765 \times {10^{23}}$ or $7.65 \times {10^{20}}$
Therefore the number of atoms of gold present in $300mg$ of gold ring of $20 - $ carat gold is $7.65 \times {10^{20}}$ .
So, the correct answer is option C) $7.6 \times {10^{20}}$ .
Note: One mole of any substance contains avogadro's number of particles .
Purity of gold is measured in parts per thousand. In order to calculate the percentage of purity of gold then we first have to divide the given carat by $24$ carat and then multiply it by $100$ . Gold less than $24 - $ carat is always an alloy that is a mixture of gold and some other metal. We cannot make jewellery from $24 - $ carat gold as it is very soft. Therefore it is mixed with other metal in order to make it hard.
Complete step by step answer:
Pure gold is $24K$ or $24$ carat.
If gold is $20$ carat this means that it contains $20$ carat of pure gold and $4$ carat is of some other metal.
So if $300mg$ of gold ring is $20 - $ carat gold.
It means that $300mg$ contains $20$ carat of pure gold and $4$ carat is of some other metal.
Substituting the values we get,
Therefore weight of pure gold in $300mg$ gold ring $ = 300 \times \dfrac{{20}}{{24}}$
Weight of pure gold in $300mg$ gold ring $ = 250mg$
Weight of pure gold in $300mg$ gold ring $ = 0.25g$
Given data :
Molecular weight of pure gold $ = 197$
To find number of moles of gold we will use the formula: $M = \dfrac{W}{{MW}}$ ,where, $M = $ Number of moles, $W = $ Weight of a substance, $MW = $ Molecular weight.
Soln:
$M = \dfrac{W}{{MW}}$
Substituting the weight of gold and molecular weight of gold we will get,
$M = \dfrac{{0.25}}{{197}}$
$M = 0.00127$
Therefore, the number of moles present in $0.25g$ of gold is $0.00127$ .
Now we will calculate number of atoms of gold present
We will use the formula given below:
Number of atoms $ = n \times {N_A}$
Where, $n = $ number of moles, ${N_A} = $ Avogadro’s number
Given data:
Number of moles of gold $ = 0.00127$
Avogadro’s number ${N_A} = 6.023 \times {10^{23}}$
Number of atoms of gold $ = n \times {N_A}$
Substituting the values in the above formula we get,
Number of atoms of gold $ = 0.00127 \times 6.023 \times {10^{23}}$
Number of atoms of gold $ = 0.00765 \times {10^{23}}$ or $7.65 \times {10^{20}}$
Therefore the number of atoms of gold present in $300mg$ of gold ring of $20 - $ carat gold is $7.65 \times {10^{20}}$ .
So, the correct answer is option C) $7.6 \times {10^{20}}$ .
Note: One mole of any substance contains avogadro's number of particles .
Purity of gold is measured in parts per thousand. In order to calculate the percentage of purity of gold then we first have to divide the given carat by $24$ carat and then multiply it by $100$ . Gold less than $24 - $ carat is always an alloy that is a mixture of gold and some other metal. We cannot make jewellery from $24 - $ carat gold as it is very soft. Therefore it is mixed with other metal in order to make it hard.
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