A 0.094 g of CuSO4*5H20is dissolved and diluted to the mark in a 500.0-mL volumetric flask. A 2.00-mL sample of this solution is transferred to a second 500.0-mL volumetric flask and diluted.
What is the molarity of CuSO4 in the final solution?
To prepare the solution directly, what mass of CuSO4*H2O would need to weighed out?
well, you have 2/500 * .094g copperIIsulfate pentahydrate in the second flask up to a volume of 500mL
molarity= molessolute/volumeinliters
answered already, 2/500*.094g
thats wrong.
To find the molarity of CuSO4 in the final solution, we need to determine the number of moles of CuSO4 in the solution.
First, we need to calculate the number of moles of CuSO4 in the initial solution. We have 0.094 g of CuSO4*5H20, which is the hydrated form of CuSO4. The molar mass of CuSO4*5H20 can be calculated as follows:
Molar mass of CuSO4 = atomic mass of Cu + atomic mass of S + 4*(atomic mass of O) = 63.55 + 32.07 + 4*(16.00) = 159.61 g/mol
Since CuSO4*5H20 contains one mole of CuSO4, we can calculate the number of moles of CuSO4 in the initial solution:
moles of CuSO4 = mass of CuSO4 / molar mass of CuSO4
= 0.094 g / 159.61 g/mol
≈ 0.0005895 mol
Next, we need to calculate the volume of the final solution. Initially, we have a 500.0 mL volumetric flask that contains the CuSO4*5H20 solution. From this, we transfer a 2.00 mL sample to a second 500.0 mL volumetric flask and dilute it. Thus, the volume of the final solution is 500.0 mL.
Finally, we can calculate the molarity of CuSO4 in the final solution using the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
Molarity of CuSO4 = moles of CuSO4 / volume of solution (in liters)
= 0.0005895 mol / (500.0 mL / 1000)
≈ 0.001179 M
Therefore, the molarity of CuSO4 in the final solution is approximately 0.001179 M.
To prepare the solution directly, i.e., without dilution, we need to calculate the mass of CuSO4*H2O that would need to be weighed out.
We know that the molar mass of CuSO4*5H20 is 159.61 g/mol. However, we need to calculate the molar mass of CuSO4*H2O, which is the anhydrous form of CuSO4. This can be obtained by subtracting the molar mass of 5H2O, which is 5*(2*(1.01) + 16.00) = 90.08 g/mol, from the molar mass of CuSO4*5H20:
Molar mass of CuSO4*H2O = molar mass of CuSO4*5H20 - molar mass of 5H2O
= 159.61 g/mol - 90.08 g/mol
= 69.53 g/mol
Now, we can calculate the mass of CuSO4*H2O that needs to be weighed out:
mass of CuSO4*H2O = molar mass of CuSO4*H2O * moles of CuSO4
= 69.53 g/mol * 0.0005895 mol
≈ 0.04099 g
Therefore, to prepare the solution directly, we would need to weigh out approximately 0.04099 g of CuSO4*H2O.