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An aqueous solution of silver nitrate (v) is electrolysed for 25 minutes using a current of 0.45a . The amount of silver deposited on the cathode is 0.755g. find the relative atomic mass of silver.
Accepted Answer
Sure, here is the calculation of the relative atomic mass of silver:
Step 1: Calculate the charge passed through the solution
Charge (Q) = Current (I) × Time (t)
=> Q = 0.45 A × 25 × 60 s
=> Q = 675 C
Step 2: Calculate the number of moles of electrons passed through the solution
No. of moles of electrons (n) = Charge (Q) / Faraday's constant (F)
=> n = 675 C / 96500 C/mol
=> n = 0.007 mol
Step 3: Calculate the number of moles of silver deposited
From the balanced chemical equation for the electrolysis of silver nitrate:
Ag+ (aq) + e- → Ag (s)
According to the equation, 1 mole of silver ion requires 1 mole of electrons for deposition.
Therefore, the number of moles of silver deposited is equal to the number of moles of electrons passed through the solution.
=> No. of moles of Ag = 0.007 mol
Step 4: Calculate the mass of silver deposited
Mass of silver (m) = No. of moles of silver (n) × Relative atomic mass of silver (Ar)
=> 0.755 g = 0.007 mol × Ar
Step 5: Calculate the relative atomic mass of silver
Relative atomic mass of silver (Ar) = Mass of silver (m) / No. of moles of silver (n)
=> Ar = 0.755 g / 0.007 mol
=> Ar = 108 g/mol
Therefore, the relative atomic mass of silver is 108 g/mol.
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