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Q.
Which of the following would contain the same number of atoms as $20\,g$ of calcium? [At. masses: $Ca=40,\,Mg=24,\,C=12$ ]
J & K CETJ & K CET 2013Some Basic Concepts of Chemistry
Solution:
$\because$ Number of atoms in $40\, g$ of
$C a=6.022 \times 10^{23}$ (Avogadro's number)
$\therefore $ Number of atoms in $20\, g$ of
$Ca =\frac{6.022 \times 10^{23} \times 20}{40}=3.011 \times 10^{23}$
Like this $\because$ Number of atoms in $24\, g$ of
$Mg =6.022 \times 10^{23}$
$\therefore $ Number of atoms in $12\, g$ of
$Mg =\frac{6.022 \times 10^{23} \times 12}{24}=3.011 \times 10^{23}$
Like this $\because$ Number of atoms in $24\, g$ of
$Mg =6.022 \times 10^{23}$
$\therefore $ Number of atoms in $12\, g$ of $Mg$
$=\frac{6.022 \times 10^{23} \times 12}{24}=3.011 \times 10^{23}$