Question

How many atoms are present in$4.25g$ ammonia?

Answer 1

Part one : Atoms

• One mole has Avogadro's number of atoms.
• The moles of given substance is given by the ratio of mass and molar mass of substance.

Part two: Calculation of atoms

• The moles of given $4.25g$ Ammonia$\left(N{H}_3\right)$ are

$$\begin{gathered} =\frac{4.25g}{17\frac{g}{mol}} \\ =0.25mol \end{gathered}$$

• The total number of atoms present in Ammonia$\left(N{H}_3\right)$ are 4.
• So , total number of atoms in $4.25g$ Ammonia$\left(N{H}_3\right)$ are

$$\begin{gathered} =4\times 0.25mol\times 6.023\times {10}^{23}\frac{1}{mol} \\ =6.023\times {10}^{23} \end{gathered}$$

Therefore, the atoms in $4.25g$ Ammonia$\left(N{H}_3\right)$ are $6.023\times {10}^{23}$