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$3.011 \times 10^{22}$ atoms of an element weight $1.15 \mathrm{~g}$. The atomic mass of the element is
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23 amu
Given, $3.011 \times 10^{22}$ atoms of an element
weight = 1.15 g
Now,
$6.023 \times 10^{23}\left(N_A\right)$ atoms of an element will weight
$$
=1.15 \times \frac{6.023 \times 10^{23}}{3.011 \times 10^{22}}=23 \mathrm{~g}
$$
The atomic mass of the element is $23 \mathrm{~g} / \mathrm{mol}$ or amu.
Note $N_A=$ number of units in one mole of any substance $=6.02214076 \times 10^{23}$.
weight = 1.15 g
Now,
$6.023 \times 10^{23}\left(N_A\right)$ atoms of an element will weight
$$
=1.15 \times \frac{6.023 \times 10^{23}}{3.011 \times 10^{22}}=23 \mathrm{~g}
$$
The atomic mass of the element is $23 \mathrm{~g} / \mathrm{mol}$ or amu.
Note $N_A=$ number of units in one mole of any substance $=6.02214076 \times 10^{23}$.
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