First slide

Faraday's Laws

Question

A 4.0 M aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to evolution of chlorine gas at one of the electrodes.
Find out the maximum weight of amalgam formed from the solution [if cathode is Hg].

Easy

A

200
B

400
C

225
D

446

Solution

no. of moles of NaCl = $\frac{M \times V (ml)}{1000} = \frac{4 \times 500}{1000} = 2 moles$
2 moles of Na formed produced two moles of Na(Hg) amalgam. Mass of amalgam $= 2 \times (23 + 200) = 446 g$

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