Question

There are $\frac{N_o}{2}$ atoms in given quantity of carbon-$12$. Same number of atoms are present in:

• A. $6g$ $Mg$
• B. $12g$ $Mg$
• C. $16g$ $O_2$
• D. $8g$ $C{H}_4$

Answer 1

The correct option is B $12g$ $Mg$

Atoms present = $\frac{N_o}{2}$
No. of moles = $\frac{1}{2}$
Mass of $\frac{1}{2}$ mol of $Mg$ = $\frac{1}{2}\times 24$ = $12g$.

Similarly,
$6$ g Mg has $N_0/4$ atoms.
$16$ g $O_2$ has $N_0$ atoms.
$8$ g $C{H}_4$ has $2N_0$ atoms.