
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
556.5k+ 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.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

