Question

How would you prepare 1.00 L of a 0.400M solution of copper(II)sulfate, CuSO4?

Answer 1

Answer: To prepare the solution of given molarity, we add 63.8 g of copper (II) sulfate to the solution.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of copper(II)sulfate solution = 0.400 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.400M=\frac{\text{Moles of copper(II)sulfate}}{1.00L}\\\\\text{Moles of copper(II)sulfate}=(0.400mol/L\times 1.00L)=0.400mol$

To calculate the mass of copper(II)sulfate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of copper(II)sulfate = 159.61 g/mol

Moles of copper(II)sulfate = 0.400 moles

Putting values in above equation, we get:

$0.400mol=\frac{\text{Mass of copper (II) sulfate}}{159.61g/mol}\\\\\text{Mass of copper (II) sulfate}=(0.400mol\times 159.61g/mol)=63.8g$

Hence, to prepare the solution of given molarity, we add 63.8 g of copper (II) sulfate to the solution.

Answer 2

We need to dilute 0.400 mol of copper (II) sulfate, how do we know, how many weigh of $CuSO_4$ we have to dilute??

It's simple.

$\eta=\frac{m}{MM}$

Using a periodic table we can find the molar mass of $Cu,~~S~~and~~O$

$MM_{CuSO_4}=153.9~g/mol$

Then

$m=\eta*MM$

now we can replace it

$m=0.400*159.6$

$\boxed{\boxed{m=63.84~g}}$

Then we have to dilute 63.84 grams of copper (II) sulfate in 1 L of water to obtain a solution with 0.400M