Hi, may I know how to solve this question? And do you have any advice for this chapter? This chapter really drives me crazy🤯 Thank you!

## Question

12.5 \mathrm{~g} of crystals of copper(II) sulphate, \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} are stirred into distilled water to make solution of concentration 1.5 \mathrm{~mol} \mathrm{dm}^{-3}. What is the volume of the solution obtained?

[Relative atomic mass: \mathrm{H}=1; \mathrm{Cu}=64, \mathrm{~S}=32, \mathrm{O}=16 ]

A 15.0 \mathrm{~cm}^{3}

B 33.3 \mathrm{~cm}^{3}

C 100 \mathrm{~cm}^{3}

D 150 \mathrm{~cm}^{3}

### Answer

Hi Jane,

I believe that your answer is correct.

The n \space\mathrm{=\frac{MV}{1000}} (where n = number of moles)

is one of the key formulas in the chapter, and I think for most questions it is important to find out the **number of moles** first, if the given information allows it.

Here, we are not given V, but we can still find the number of moles of \mathrm{CuSO_4} used because the question gives us information regarding **relative atomic mass**. By adding the given relative atomic masses to find the **relative formula mass**, we can then find the number of moles.

number of moles can be found by using: mass used \div relative formula mass

For making stock solution (whether by dissolving solid or by diluting liquid), the number of moles of the solid being dissolved / liquid being diluted doesn’t change. (because we are not pouring away the solid or liquid, only adding more distilled water to it)

Only the concentration changes: concentration decreases as you add more volume, obeying the logic of the formula n \space\mathrm{=\frac{MV}{1000}}.

Maybe more topical practices of reference book exercises / state modules / state trial papers will help, and really try to understand what is actually happening and why the formula works. Hope this helps!

Okie! Thank you so much☺️