Sodium Electron Number Density Calculation

This solution determines the number density of valence electrons in Sodium (Na) by utilizing its given atomic mass, mass density, and Avogadro's number.

Key Information Provided

• Atomic Mass of Na ($M$): $23$ g mol$^{-1}$
• Mass Density of Na ($\rho$): $0.968$ g cm$^{-3}$
• Avogadro Number ($N_A$): $6.022 \times 10^{23}$ mol$^{-1}$
• Valence electrons per Sodium atom ($Z$): $1$ (Sodium is an alkali metal)

Calculation Formula

The number density of valence electrons ($n_e$) per unit volume is calculated using the formula:

$n_e = \frac{\text{Density}}{\text{Atomic Mass}} \times \text{Avogadro Number} \times \text{Valence Electrons per Atom}$

Which simplifies to:

$n_e = \frac{\rho}{M} \times N_A \times Z$

Step-by-Step Calculation

1. Calculate the number density of atoms ($n_{atoms}$):

   $n_{atoms} = \frac{\rho}{M} \times N_A = \frac{0.968 \text{ g cm}^{-3}}{23 \text{ g mol}^{-1}} \times (6.022 \times 10^{23} \text{ mol}^{-1})$

   $n_{atoms} \approx 0.042087 \times 6.022 \times 10^{23}$ cm$^{-3}$

   $n_{atoms} \approx 0.25345 \times 10^{23}$ atoms cm$^{-3}$

2. Calculate the number density of valence electrons ($n_e$):

   Since each Sodium atom has $1$ valence electron ($Z=1$):

   $n_e = n_{atoms} \times Z = (0.25345 \times 10^{23} \text{ cm}^{-3}) \times 1$

   $n_e \approx 2.5345 \times 10^{22}$ electrons cm$^{-3}$

3. Format the result:

   The question asks for the value in the format $\_\_\_\_\times 10^{22}$ cm$^{-3}$, rounded to two decimal places.

   Rounding $2.5345$ to two decimal places gives $2.53$.

Final Answer

The number density of valence electrons is approximately $2.53$ $\times 10^{22}$ cm$^{-3}$.