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One atom of an element x weight $6.643\times {{10}^{-23}}g$ number of moles of atom in $20kg$ is​:
A.$4$
B.$40$
C.$100$
D.$500$

Answer
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Hint: We know that on multiplying molar mass of substance to the number of molecules provided in the problem we get grams of substance contained in a particular mole of a substance. Atomic weight when multiplied to the number of mole results in its gram weight. Here we will use two formulas
1.\[molar\text{ }mass\text{ }=\text{ }mass\text{ }of\text{ }{{N}_{A}}\text{ }atoms\]
2.\[Grams\text{ a}toms~=~\dfrac{Mass}{Molar\text{ }Mass}\]

Complete step-by-step answer:Here mole is standard unit of measurement of amount of substance. Substances react in simple ratios of their respective mole. One mole consists of Avogadro number of atoms that is $6.022\times {{10}^{23}}atom$ . Thus the molar mass is the amount of gram present in one mole of substance. It's also the average atomic weight of a substance . For example , carbon has an atomic weight of $12amu$ so a molar mass is given as $12$ gram per mole. The formula for conversion of molar mass of atom to gram is given as:
Mass of $1$ atom $=6.644\times {{10}^{-23}}g$
Similarly, \[molar\text{ }mass\text{ }=\text{ }mass\text{ }of\text{ }{{N}_{A}}\text{ }atoms\]
\[molar\text{ }mass\text{ }=\left( 6.022\times {{10}^{23}} \right)\times \left( 6.644\times {{10}^{-23}} \right)=40.01\dfrac{g}{mol}\]
Thus we get molar mass is $40.01\text{ }g\cdot mo{{l}^{-1}}$ . Now that we get molar mass and grams atoms is given by,
 \[Grams\text{ a}toms~=~\dfrac{Mass}{Molar\text{ }Mass}~=\dfrac{20\times {{10}^{3}}}{40.01\text{ }g\cdot mo{{l}^{-1}}}=500\text{ }gram\text{ atom}\]
Therefore, Option D is the correct option i.e. one atom of an element x weight $6.643\times {{10}^{-23}}g$ number of moles of atom in $20kg$ is 500 gram atom.

Note: Note that calculating mass is known as gram atomic mass. It's mass of one mole of element example; gram atomic mass of a hydrogen which is $1$ gram for one mole of hydrogen. It's the same as the atomic weight of a hydrogen atom.