Number of atoms present in (4.25 ,g) of (NH\_3 ) is | Chemistry | SolveeIt

Number of atoms present in $4.25 \,g$ of $NH_3$ is

$6.023\times 10^{23}$

$4\times 6.023\times 10^{23}$

$1.7\times 10^{24}$

$4.25\times 6.023\times 10^{23}$

Answer

$6.023\times 10^{23}$

Explanation

Number of atoms in $4.25 \,g$ of $NH_3$ $= \frac{4.25}{17} \times N_A \times 4$ $= N_A = 6.032 \times 10^{23}$

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