How much will $1.2 \times {10^{46}}$ atoms of carbon weigh?

Hint: We know that the gram atomic mass of any element contains $6.022 \times {10^{23}}$atoms in it, this number is what we call Avogadro’s number and is represented by the symbol ${N_0}$. Avogadro’s number is the very same number which represents the amount of particles present in$1$mole of a substance irrespective of whether the particle is an atom, molecule, element or compound. So, one mole of carbon atoms is equal to $6.022 \times {10^{23}}$ carbon atoms. To solve such questions we need a proper knowledge of unitary methods.

Complete step-by-step answer:
We all know that the gram atomic mass of carbon is equal to $12$ grams.
Thus by using unitary method,
$6.022 \times {10^{23}}$Atoms of carbon weigh$ = $$12$ gram
 $1$ Atom of carbon would weigh$ = \dfrac{{12}}{{6.022 \times {{10}^{23}}}}$ gram
$1.2 \times {10^{46}}$Atoms of carbon would weigh $ = \dfrac{{12}}{{6.022 \times {{10}^{23}}}} \times 1.2 \times {10^{46}}$$ = 2.4 \times {10^{23}}g$
Hence, $ = 2.4 \times {10^{23}}g$ is the required answer.

Additional information:
$1$ Gram atomic mass or one-gram atom is the mass of $\dfrac{1}{{12}}th$ the mass of$C - 12$ atoms. 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.