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Sagot :
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
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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