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Q.
$2.0\, g$ of oxygen contains number of atoms equal to that in:
J & K CETJ & K CET 2000
Solution:
Number of atoms in $1\, g$ of an element $=\frac{6.023 \times 10^{23}}{\text { atomic mass }}$
$\therefore $ Number of atoms in $2\, g$ of oxygen $=\frac{6.023 \times 10^{23} \times 2}{16}$
$=\frac{6.023 \times 10^{23}}{8}$
and number of atoms in $4 \,g$ of sulphur
$=\frac{6.023 \times 10^{23} \times 4}{32}$
$=\frac{6.023 \times 10^{23}}{8}$