Huh ion speak that language:/
Answer : The original concentration of copper (II) sulfate in the sample is, ![5.6\times 10^{-1}g/L](https://tex.z-dn.net/?f=5.6%5Ctimes%2010%5E%7B-1%7Dg%2FL)
Explanation :
Molar mass of Cu = 63.5 g/mol
First we have to calculate the number of moles of Cu.
Number of moles of Cu = ![\frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}=\frac{89\times 10^{-3}g}{63.5g/mol}=1.40\times 10^{-3}mole](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20Cu%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20Cu%7D%7D%3D%5Cfrac%7B89%5Ctimes%2010%5E%7B-3%7Dg%7D%7B63.5g%2Fmol%7D%3D1.40%5Ctimes%2010%5E%7B-3%7Dmole)
Now we have to calculate the number of moles of ![CuSO_4](https://tex.z-dn.net/?f=CuSO_4)
Number of moles of Cu = Number of moles of ![CuSO_4](https://tex.z-dn.net/?f=CuSO_4)
Number of moles of
= ![1.40\times 10^{-3}mole](https://tex.z-dn.net/?f=1.40%5Ctimes%2010%5E%7B-3%7Dmole)
Now we have to calculate the molarity of ![CuSO_4](https://tex.z-dn.net/?f=CuSO_4)
![\text{Molarity}=\frac{\text{Moles of }CuSO_4\times 1000}{\text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DCuSO_4%5Ctimes%201000%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
Now put all the given values in this formula, we get:
![\text{Molarity}=\frac{1.40\times 10^{-3}mole\times 1000}{400.mL}=0.0035M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%7D%3D%5Cfrac%7B1.40%5Ctimes%2010%5E%7B-3%7Dmole%5Ctimes%201000%7D%7B400.mL%7D%3D0.0035M)
To change mol/L into g/L, we need to multiply it with molar mass of ![CuSO_4](https://tex.z-dn.net/?f=CuSO_4)
Molar mass of
= 159.609 g/mL
Concentration in g/L = ![0.0035M\times 159.609g/mol=0.5586g/L\approx 5.6\times 10^{-1}g/L](https://tex.z-dn.net/?f=0.0035M%5Ctimes%20159.609g%2Fmol%3D0.5586g%2FL%5Capprox%205.6%5Ctimes%2010%5E%7B-1%7Dg%2FL)
Thus, the original concentration of copper (II) sulfate in the sample is, ![5.6\times 10^{-1}g/L](https://tex.z-dn.net/?f=5.6%5Ctimes%2010%5E%7B-1%7Dg%2FL)
DEEZ NUTS ha gattti -sorry