Problem 3

A volume of 200 cm

^3

of a 2-normal sodium chloride solution (

\rho = 1.10

g cm

^{-3}

) was electrolysed at permanent stirring in an electrolytic cell with copper electrodes. Electrolysis was stopped when 22.4 dm

^3

(at STP) of a gas were liberated at the cathode. Calculate the mass percentage of NaCl in the solution after electrolysis. Relative atomic masses:

A_r(\text{H}) = 1

;

A_r(\text{O}) = 16

;

A_r(\text{Na}) = 23

;

A_r(\text{Cl}) = 35.5

;

A_r(\text{Cu}) = 64

Step 3 of 3: Final mass percentage

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m(\text{NaCl}) = 2 \times 0.200 \times 58.5 = 23.4\ \text{g}; \quad m(\text{solution}) = 200 \times 1.10 - 36 = 184\ \text{g}; \quad w = \dfrac{23.4}{184} = 12.7\ \%

Analysis

Initial NaCl: $0.2 \times 2 = 0.4$ mol = 23.4 g (unchanged through the cycle). Initial solution mass $200 \times 1.10 = 220$ g; minus 36 g of decomposed water leaves 184 g, so $w(\text{NaCl}) = 23.4/184 = 12.7$ %.

Common pitfall. The trap is to assume NaCl is consumed; with copper electrodes it is regenerated and only the solvent mass drops.