Answer
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Hint: We should be quite familiar with the unitary method in order to solve these types of questions. It may be noted that the gram atomic mass of any element has $6.022 \times {10^{23}}$ atoms. We should also note that a mole represents Avogadro’s number or $6.022 \times {10^{23}}$ particles. ${N_0}$ or Avogadro’s number is the very same number which represents the number of particles present in $1$ mole of a substance irrespective of whether the particle is an atom, molecule, element or compound.
One mole of lead atoms is equal to $6.022 \times {10^{23}}$lead atoms.
Complete step by step solution:
We all know that the gram atomic mass of lead is equal to $207.2$ grams.
By using unitary method,
$207.2$ grams of lead would contain $6.022 \times {10^{23}}$ lead atoms
$1\,$ gram of lead would contain $\dfrac{{6.022 \times {{10}^{23}}}}{{207.2}}$ lead atoms
$406.4\,g$ of lead would contain $\dfrac{{6.022 \times {{10}^{23}}}}{{207.2}} \times 406.4$ lead atoms $ = 1.18 \times {10^{24}}$ lead atoms.
Hence, $1.18 \times {10^{24}}$ is the required answer.
Additional information: Gram atomic mass or one-gram atoms is the mass of $6.022 \times {10^{23}}$ atoms (or one mole atoms) of an element expressed in grams.
Molar mass is the mass of $6.022 \times {10^{23}}$ atoms (or one mole atom) of an element expressed in grams.
Note: Here precaution must be taken while handling the numerical calculations as a small error in calculation could make a drastic change thus, we must try to do calculations that are precise to the very decimal points. Also, we must take note about the units of weight and molar mass in the question as this sometimes leads to grave calculative errors especially when weight is given in units other than the standard S.I. units of $grams\left( {gm} \right)$, commonly the weight is provided in units of microgram, milligram, nanogram etc. other than the standard units of gram. Another common mistake done by students is not reading what is asked in the question properly; sometimes we are tasked with finding the number of oxygen atoms in moles and not in grams. In those cases we should only provide answers to what is asked and not to what all we can.
One mole of lead atoms is equal to $6.022 \times {10^{23}}$lead atoms.
Complete step by step solution:
We all know that the gram atomic mass of lead is equal to $207.2$ grams.
By using unitary method,
$207.2$ grams of lead would contain $6.022 \times {10^{23}}$ lead atoms
$1\,$ gram of lead would contain $\dfrac{{6.022 \times {{10}^{23}}}}{{207.2}}$ lead atoms
$406.4\,g$ of lead would contain $\dfrac{{6.022 \times {{10}^{23}}}}{{207.2}} \times 406.4$ lead atoms $ = 1.18 \times {10^{24}}$ lead atoms.
Hence, $1.18 \times {10^{24}}$ is the required answer.
Additional information: Gram atomic mass or one-gram atoms is the mass of $6.022 \times {10^{23}}$ atoms (or one mole atoms) of an element expressed in grams.
Molar mass is the mass of $6.022 \times {10^{23}}$ atoms (or one mole atom) of an element expressed in grams.
Note: Here precaution must be taken while handling the numerical calculations as a small error in calculation could make a drastic change thus, we must try to do calculations that are precise to the very decimal points. Also, we must take note about the units of weight and molar mass in the question as this sometimes leads to grave calculative errors especially when weight is given in units other than the standard S.I. units of $grams\left( {gm} \right)$, commonly the weight is provided in units of microgram, milligram, nanogram etc. other than the standard units of gram. Another common mistake done by students is not reading what is asked in the question properly; sometimes we are tasked with finding the number of oxygen atoms in moles and not in grams. In those cases we should only provide answers to what is asked and not to what all we can.
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