Question

Find out the total valence electron in 360 mg of $NH_4^+$ ion sample.

Answer 1

$9.6 \times 10^{22}$

Answer 2

To find the total valence electrons in 360 mg of $NH_4^+$ ion sample, follow these steps:

1. Calculate the number of valence electrons in one $NH_4^+$ ion:

   • Nitrogen (N) is in Group 15, so it has 5 valence electrons.
   • Hydrogen (H) is in Group 1, so it has 1 valence electron.
   • In a neutral $NH_4$ molecule (hypothetical), the total valence electrons would be: 5 (from N) + 4 × 1 (from 4 H atoms) = 9 valence electrons.
   • The ion is $NH_4^+$, which means it has lost one electron. This lost electron is a valence electron.
   • Therefore, the number of valence electrons in one $NH_4^+$ ion = 9 - 1 = 8 valence electrons.

2. Calculate the molar mass of $NH_4^+$:

   • Atomic mass of Nitrogen (N) = 14 g/mol
   • Atomic mass of Hydrogen (H) = 1 g/mol
   • Molar mass of $NH_4^+$ = (1 × 14) + (4 × 1) = 14 + 4 = 18 g/mol.

3. Convert the given mass of the sample to grams:

   • Given mass = 360 mg = 360 × $10^{-3}$ g = 0.360 g.

4. Calculate the number of moles of $NH_4^+$ ions in the sample:

   • Number of moles = $\frac{\text{Mass}}{\text{Molar mass}}$
   • Number of moles = $\frac{0.360 \text{ g}}{18 \text{ g/mol}} = 0.02 \text{ mol}$.

5. Calculate the total number of $NH_4^+$ ions in the sample:

   • Use Avogadro's number ($N_A = 6.022 \times 10^{23}$ ions/mol). For calculation simplicity, often $N_A = 6 \times 10^{23}$ is used in exams if not specified. We will use $6 \times 10^{23}$.
   • Number of ions = Number of moles × Avogadro's number
   • Number of ions = $0.02 \text{ mol} \times 6 \times 10^{23} \text{ ions/mol}$
   • Number of ions = $0.12 \times 10^{23} \text{ ions} = 1.2 \times 10^{22} \text{ ions}$.

6. Calculate the total valence electrons in the sample:

   • Total valence electrons = Number of $NH_4^+$ ions × Valence electrons per $NH_4^+$ ion
   • Total valence electrons = $1.2 \times 10^{22} \text{ ions} \times 8 \text{ valence electrons/ion}$
   • Total valence electrons = $9.6 \times 10^{22}$ valence electrons.