Question

(7 pts) Copper sulfate is used as a dietary supplement for animal feed. A technician prepares a "stock solution of CuSO4 by adding 30.00 g of CuSO, to water. After the solid is dissolved the technician dilutes the solution to a final volume of 500.0 ml. A. Calculate the Molarity of the CuSO,"stock” solution prepared by the technician. B. Describe in a complete sentence how the technician would prepare 2.50 L of 0.100 M CuSO. solution using the "stock" CuSO, solution. Be sure to show any calculations.

Answer 1

7.A We know that

Molarity (M)=moles of solute(n)/Volume of solution(L)

Moles of solute (n)= mass of solute(m)/molar mass of solute(MW)

Given,

Mass of CuSO4 added = 30.00 g

Volume of solution after dilution = 500.0 mL = 0.5 L

(1L=1000 mL)

Molar mass of CuSO4= 159.609 g/mol

Moles of CuSO4 (n) = Mass of CuSO4(m)/Molar mass of CuSO4(MW)

Moles of CuSO4(n) = (30.00 g)/(159.609 g/mol)=0.188 mol

Molarity (M) = moles of CuSO4(n)/Volume of solution(L)

Molarity(M) = 0.188 mol/0.5 L = 0.376 mol/L = 0.376 M

Hence, the molarity of CuSO4 stock is 0.376 M

7.B As the molarity of the solution to be prepared is lower than that of stock solution. So, using the dilution formula, we can calculate the amount of stock solution to be taken.

M1V1=M2V2 .......(1) dilution formula

M1=Molarity of stock solution =0.376 M

V1=Volume of stock solution = ?

M2=Molarity of solution to be prepared =0.100 M

V2=Volume of solution to be prepared = 2.50 L

Using (1), we get

0.376 M x V1 = 0.100 M x 2.50 L

=> V1 = (0.100 M x 2.50 L)/(0.376 M) = 0.665 L = 665 mL

To prepare the 2.50 L of 0.100 M CuSO4, the technician needs 665 mL stock CuSO4 solution. But the technician prepared only 500 mL stock solution. He needs to prepare more stock solution.