Question

The total number of atoms present in 34 g of $N{H_3}$ is:
A. $4 \times {10^{23}}$
B. $4.8 \times {10^{21}}$
C. $2 \times {10^{23}}$
D. $48 \times {10^{23}}$

Answer 1

Hint: To calculate the total number of atoms first we need to calculate the molar mass of the compound. The one mol of substance contain $6.022 \times {10^{23}}$ molecules (Avagadro’s number) molecule so the number of molecules is calculated by dividing mass by molecular mass multiplied by the Avagadro’s number.

Complete step by step answer:
The mass of $N{H_3}$ is 34 g.
The molar mass is calculated by adding the atomic weight of the atoms and multiplying the number of atoms with it.
The mass of 1 mol ($6.022 \times {10^{23}}$) of $N{H_3}$ is calculated as shown below.
$\Rightarrow$ Mass of $N{H_3}$= $14 + 3 \times 1$
$\Rightarrow$ Mass of $N{H_3}$= 17 g.
The number of molecules is calculated by dividing mass by molecular mass multiplied by avagadro’s number.
The Avagadro’s number is the unit in one mole of any substance also stated as molecular weight in grams equal to $6.022 \times {10^{23}}$.
The number of molecules in 34 g $N{H_3}$ is shown below.
$\Rightarrow \dfrac{{34}}{{17}} \times 6.022 \times {10^{23}}$
$\Rightarrow 12.044 \times {10^{23}}$
In $N{H_3}$ molecule one nitrogen atom and three hydrogen atoms are present.
So in 1 mol of $N{H_3}$ molecule, the total number of atoms present is 4.
$\Rightarrow$The number of atoms present in 34 g of $N{H_3}$= $4 \times 12.044 \times {10^{23}}$
$\Rightarrow$The number of atoms present in 34 g of $N{H_3}$= $48 \times {10^{23}}$
Thus, the number of atoms present in 34 grams of $N{H_3}$ is $48 \times {10^{23}}$.

So, the correct answer is Option D.

Note: Make sure that 1 mol of $N{H_3}$ molecule contains 4 atoms and we need to find out the number of atoms present in 34 g of $N{H_3}$. The property of Avagadro’s number is that the mass of 1 mole substance is the same as the molar mass of the substance.