Calculate the cost to deposit 24 g of Copper, given that the cost to deposit 1 g of Aluminium is N40, M(Al)=27 g/mol, M(Cu)=64 g/mol, valency of Al=3, and valency of Cu=2.

The equivalent mass ($E$) is given by the formula: $E = \frac{\text{Molar mass}}{\text{Valency}}$.

For Aluminium:

$$E_{Al} = \frac{M_{Al}}{n_{Al}} = \frac{27 \text{ g/mol}}{3} = 9 \text{ g/equivalent}$$

For Copper:

$$E_{Cu} = \frac{M_{Cu}}{n_{Cu}} = \frac{64 \text{ g/mol}}{2} = 32 \text{ g/equivalent}$$

We are given that it costs N40 to deposit 1 g of Aluminium.
First, find how many equivalents are in 1 g of Aluminium:

$$\text{Equivalents of Al} = \frac{\text{Mass of Al}}{E_{Al}} = \frac{1 \text{ g}}{9 \text{ g/equivalent}} = \frac{1}{9} \text{ equivalents}$$

Now, we can find the cost to deposit 1 equivalent of Aluminium:

$$\text{Cost per equivalent} = \frac{\text{Cost for 1 g Al}}{\text{Equivalents of Al in 1 g}} = \frac{\text{N}40}{\frac{1}{9} \text{ equivalents}} = \text{N}40 \times 9 = \text{N}360 \text{ per equivalent}$$

According to Faraday's second law, the same quantity of electricity (and thus the same cost) will deposit one equivalent of any substance. Therefore, the cost per equivalent for Copper will also be N360.

First, find how many equivalents are in 24 g of Copper:

$$\text{Equivalents of Cu} = \frac{\text{Mass of Cu}}{E_{Cu}} = \frac{24 \text{ g}}{32 \text{ g/equivalent}} = \frac{3}{4} \text{ equivalents}$$

Now, calculate the total cost to deposit 24 g of Copper:

$$\text{Cost for Cu} = \text{Equivalents of Cu} \times \text{Cost per equivalent}$$ $$\text{Cost for Cu} = \frac{3}{4} \times \text{N}360$$ $$\text{Cost for Cu} = 3 \times \text{N}90$$ $$\text{Cost for Cu} = \text{N}270$$

Comparing this to the given options, the correct answer is N270.00.

The final answer is $\text{N}270.00$.