Question

Do not understand how to solve questions 2 & 3 for the lab.

Answer 1

Answer:

Molar mass of CuSO₄.5H₂O = 249.68 g.mol^-1.

2. We are asked to prepare 40.0 mL of 0.1 M CuSO₄.5H₂O.

Molarity = 0.1 M = 0.1 mol/L and Volume = 40.0 mL = 0.040 L

We know that, we calculate # of moles of substance in solution as,

# of moles = Molarity of solution * Volume of solution in L.

$\therefore$ # of moles of CuSO₄.5H₂O = 0.1 mol/L * 0.040 L = 0.004 mol.

By knowing # of moles of substance we can ind its mass in solution as,

Mass of substance = # of moles of substance * Molar mass of substance

Mass of CuSO₄.5H₂O = 0.004 mol * 249.68 g.mol^-1 = 0.999 g

0.999 g i.e. 999 mg of CuSO₄.5H₂O dissolved and diluted to 40.0 mL of distilled water gives 0.1 M CuSO₄.5H₂O solution in 40.0 mL distilled water.

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2) This problem can be solved using Law of dilution stated as,

M1V1 = M2V2. ............ (1)

Where, M1, V1 are molarity and volume of original or say stock solution and M2, V2 that of new solution to be prepared.

For our problem,

M1 = 0.1 M and V1 = ?

M2 = 0.06 M, V2 = 10.0 mL

Placing these values in eq.(1)

0.1 M * V1 = 0.06 M * 10.0 mL

V1 = 6.0 mL

6.0 mL of 0.1 M CuSO₄.5H₂O stock when diluted to 10 mL will give 0.06 M CuSO₄.5H₂O solution.

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