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$3.011 \times 10^{22}$ atoms of an element weighs $1.15 \mathrm{~g}$. The atomic mass of the element is
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The correct answer is:
$23$
From Avogadro's law,
Avogadro's number $=$ Atomic mass of the element
$\because 3.011 \times 10^{22}$ atoms of an element weigh $=1.15 \mathrm{~g}$
$\therefore 6.023 \times 10^{23}$ atoms weigh
$=\frac{1.15 \times 6.023 \times 10^{23}}{3.011 \times 10^{22}}=23$
Avogadro's number $=$ Atomic mass of the element
$\because 3.011 \times 10^{22}$ atoms of an element weigh $=1.15 \mathrm{~g}$
$\therefore 6.023 \times 10^{23}$ atoms weigh
$=\frac{1.15 \times 6.023 \times 10^{23}}{3.011 \times 10^{22}}=23$
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