Answer:
4.61 x 10ˉ⁴ mole H₂(g)
Explanation:
Given electric current for 2 min. => 12.4ml of wet H₂(g) at 25⁰C & 715Torr => ? moles H₂(g)
PV = nRT => n = PV/RT
• P = P(H₂) + P(H₂O) = 715Torr = 715mm = P(H₂) + 23.8mm
=> P(H₂) = (715 – 23.8)mm = 619.2mm = 619.2mm/760mm/Atm = 0.815Atm
• V = 12.8ml = 0.0128L
• R = 0.08206 L∙Atm/mol∙K
• T = 25⁰C = 298K
Substituting … n = PV/RT
=> n = (0.815Atm)(0.0128L)/(0.08206 L∙Atm/mol∙K)(298K) = 4.61 x 10ˉ⁴ mole H₂(g)