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

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

It's simple.

[tex]\eta=\frac{m}{MM}[/tex]

Using a periodic table we can find the molar mass of [tex]Cu,~~S~~and~~O[/tex]

[tex]MM_{CuSO_4}=153.9~g/mol[/tex]

Then

[tex]m=\eta*MM[/tex]

now we can replace it

[tex]m=0.400*159.6[/tex]

[tex]\boxed{\boxed{m=63.84~g}}[/tex]

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

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:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Molarity of copper(II)sulfate solution = 0.400 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

[tex]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[/tex]

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

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:

[tex]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[/tex]

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

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