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# The number of atoms in 4.25g of $NH_3$ is approximately

$\begin{array}{1 1}(a)\;1\times 10^{23}&(b)\;1.5\times 10^{23}\\(c)\;2\times 10^{23}& (d)\;6\times 10^{23}\end{array}$

17gm of $NH_3=6.023\times 10^{23}$ mole
$\therefore$ 4.25g of $NH_3=\large\frac{6.023\times 10^{23}}{17}$$\times 4.25 \therefore No of atoms in NH_3=\large\frac{4\times 6.023\times 10^{23}}{17}$$\times 4.25\approx 6\times 10^{23}$
$\therefore$ The option is (d) $6\times 10^{23}$ .