JAMB Chemistry 2019 Paper

X$^{3+}$ + 3e$^{-}$ $\rightarrow$ X

3F = 27g

xF = 10g

$\frac{x}{3} = \frac{10}{27} \implies x = \frac{10}{9} F$

$\frac{10}{9}$F $\equiv$ N20.00

1F is equivalent to x

$\frac{1}{\frac{10}{9}} = \frac{x}{20}$

$\frac{9}{10} = \frac{x}{20} \implies x = N18.00$

1F is equivalent to N18.00.

Y$^{2+}$ + 2e$^{-}$ $\rightarrow$ Y

2F = 24g

xF = 6g

x = $\frac{6 \times 2}{24} = \frac{1}{2} F$

1F = N18.00

$\frac{1}{2}$F = $\frac{1}{2} \times N18.00$

= N9.00