Answer:
[tex]\boxed{6.08 \times 10^{23}}[/tex]
Explanation:
Data:
I = 2.15 A
t = 8 min 24 s
T = 26.0 °C
V = 65.4 mL
p = 774.2 To
1. Write the equation for the half-reaction
2H₂O ⟶ O₂ + 4H⁺ + 4e⁻
2. Calculate the moles of oxygen
[tex]p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}[/tex]
V = 0.0654 L
T = (26.0 + 273.15) K = 299.15 K
[tex]\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}[/tex]
3. Calculate the moles of electrons
[tex]\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}[/tex]
4. Calculate the number of coulombs
t = 8 min 24 s =504 s
Q = It = 504 s × 2.10 C·s⁻¹= 1058 C
5. Calculate the number of electrons
[tex]\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}[/tex]
6. Calculate Avogadro's number
[tex]N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}[/tex]
