JAMB Chemistry 2019 Paper

Concept:Faraday’s laws of electrolysis connect the mass discharged with the charge passed. Cost is proportional to the charge, so first find the cost per faraday.Explanation:For $X^{3+}$, the reduction equation is$$X^{3+} + 3e^{-} \rightarrow X$$Here, $3\text{F}$ discharges $27\text{g}$ of $X$.If the charge needed to discharge $10\text{g}$ is $Q$ faradays, then$$\frac{Q}{3} = \frac{10}{27}$$$$Q = \frac{30}{27}\text{F} = \frac{10}{9}\text{F}$$This charge costs $\text{N}20.00$, so $1\text{F}$ costs$$\frac{20}{\frac{10}{9}} = 20 \times \frac{9}{10} = \text{N}18.00$$For $Y^{2+}$, the reduction is$$Y^{2+} + 2e^{-} \rightarrow Y$$$2\text{F}$ discharges $24\text{g}$ of $Y$.Let the charge for $6\text{g}$ be $Q$ faradays.$$\frac{Q}{2} = \frac{6}{24}$$$$Q = \frac{12}{24}\text{F} = \frac{1}{2}\text{F}$$Cost of $\frac{1}{2}\text{F} = \frac{1}{2} \times \text{N}18.00 = \text{N}9.00$Answer:D. N9.00