(3) if an aqueous solution of hno3 is electrolyzed for 34.00 min at a steady current of 1.83 a, what volume of h2 (g) at 25.0°c and 1.05 atm will be collected at the cathode?

When 2H+  + 2e-  →  H2(g)

when the total charge transfer = steady current * time
                                                    =1.83 * 34 min * 60s = 3733.2 C
and when each electron has 1.6022x10^-19 C
So the amount of charge = 3733.2 / 1.6022x10^-19
                                           = 2.33x10^22 electrons 
To get the moles of the electrons we divided by Avogadro's number
= 2.33x10^22 / 6.022x10^23 = 0.03869 moles
when 2mol e → 1 mol H2
So      0.03869→ ???
moles of H2 = 0.03869 /2 = 0.01935 moles
know we have n = 0.01935 & p = 1.05 atm & R= 0.0821 & T = 25+273=298 K

So by substitution in the ideal gas formula, we can get V:

PV = nRT
1.05 * V = 0.01935 * 0.0821 * 298
∴ V  = 0.45 L

I’m really not sure what’s the answer. sorry