Question

In a similar experiment, a current of 2.15 amps ran for 8 minutes and 24 seconds. The temperature of the water was 26.0°C. The volume of O2 collected was 65.4 mL. The pressure, after all corrections have been made, was 774.2 to. Calculate Avogadro's number using these data.

Answer 1

`\boxed{6.08 * 10^(23)}`

Step-by-step 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

`p = \text{774.2 To} * \frac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}`

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

`\begin{array}{rcl}pV& = & nRT\\1.0189 * 0.0654 & = & n * 0.08206 * 299.15\\0.06662 & = & 24.55n\\\\n & = & (0.06662)/(24.55)\\\\n & = & 2.714 * 10^(-3)\\\end{array}`

3. Calculate the moles of electrons

`\text{n} = \text{2.714 $* 10^(-3)$ mol oxygen} * \frac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}`

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

`\text{No. of electrons} = \text{1058 C} * \frac{\text{1 electron}}{1.602 * 10^(-19)\text{ C}} = 6.607 * 10^(21)\text{ electrons}`

6. Calculate Avogadro's number

`N_{\text{A}} = (6.607 * 10^(21))/(0.01086) = \mathbf{6.08 * 10^(23)}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 * 10^(23)}}`