If 13.8 , g of an element contains 4.6 times | vedclass

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(B) The number of atoms in $1 \, mole$ of an element is equal to Avogadro's number,$N_A = 6.022 	imes 10^{23} \, atoms/mol$. Given that $4.6 	imes 1

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$180$

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(B) The number of atoms in $1 \, mole$ of an element is equal to Avogadro's number,$N_A = 6.022 	imes 10^{23} \, atoms/mol$. Given that $4.6 	imes 10^{22}$ atoms weigh $13.8 \, g$. Therefore,the mass of $6.022 	imes 10^{23}$ atoms (atomic mass) is calculated as: $	ext{Atomic Mass} = \frac{13.8 \, g 	imes 6.022 	imes 10^{23} \, atoms/mol}{4.6 	imes 10^{22} \, atoms} = 180 \, g/mol$.

If $13.8 \, g$ of an element contains $4.6 \times 10^{22}$ atoms,what is the atomic mass of that element?

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