Question

How many grams of $CuS{O_4}.5{H_2}O$ are needed to prepare $100ml$ of a $0.10M$ solution?

Answer 1

$CuS{O_4}.5{H_2}O$ , an ionic compound that contains water of crystallization in its structure. Generally, you have five moles of water of crystallization for everyone mole of anhydrous copper(II) sulfate.
This means that you're going to have to account for the mass of this water of crystallization in your calculations. Now, you need your target solution to have a molarity of $0.10M$ and a volume of $100ml$ .

Complete step by step answer:
Now, you need your target solution to have a molarity of $0.10M$ and a volume of $100ml$ .
$C = \dfrac{n}{V}$
Where, $C$ = Concentration (molarity)
$n$ = number of moles
$V$ = volume of solution
From the above question, concentration = $0.10M$ and volume of solution = $100ml$ . So, by substituting the values we will determine the number of moles.
${n_{(CuS{O_4})}} = 0.10{L^{ - 1}} \times 100 \times {10^{ - 3}}L$
${n_{(CuS{O_4})}} = 0.0100moles$
Now you have to determine how many grams of copper(II) sulfate pentahydrate would contain this many grams of copper(II) sulfate.
To do that, calculate the hydrate's percent composition by using its molar mass and the molar mass of anhydrous copper(II) sulfate.
Remember that there are 5molesof water for every 1mole of hydrate
$\dfrac{{159.61gmo{l^{ - 1}}}}{{(159.61 + 5 \times 18.015)gmo{l^{ - 1}}}} \times 100\% = 63.92\% CuS{O_4}$
This tells you that for every $100g$ of copper(II) sulfate pentahydrate, you get $63.92\%$ of anhydrous copper(II) sulfate.
$1.596gCuS{O_4} \times \dfrac{{100gCuS{O_4}.5{H_2}O}}{{63.92gCuS{O_4}}} = 2.5gCuS{O_4}.5{H_2}O$

Note: It's important to realize that your solute is anhydrous copper(II) sulfate, $CuS{O_4}$ . Use the anhydrous salt's molar mass to determine how many grams of copper(II) sulfate would contain that many moles. Make sure that you are reading the ques carefully and understanding what is asked.