You will be given a substance and concentration for the solution. The substance

Question

You will be given a substance and concentration for the solution. The substance will be a salt. After the instructor verifies that your calculations are correct, you will prepare a stock solution of this substance and, from that stock solution, dilute a portion to make a less concentrated solution. Upon creating your diluted solution, you will bring your sample to the instructor. The instructor will test your solution's concentration. If your solution is too far off from the expected value, then you will need to start over. Volumetric flasks and pipet must be used to measure volume of solutions. The salt for today is CuCl_2. 2H_2O. You will make 50.00 mL of approximately 0.15 M stock solution and 25.00 mL of 0.045 M diluted solution.

Explanation / Answer

To make 50 ml ( 0.05 L ) of 0.15 M ( mol/L) stock solution

No. of moles of salt required = 0.05 L * 0.15 mol/L = 0.0075 moles of salt

Molar mass of salt (CuCl2.2H2O) = 170.5 g/mol

Mass of salt required = Molar mass * No. of moles = 170.5 * 0.0075 = 1.28 g of salt (CuCl2.2H2O)

Approximately 1.28 g of salt (CuCl2.2H2O) should be dissolved in 50 ml of water to make stock solution

to dilute to 25 ml (0.025 L) of 0.045 M of solution from stock solution M1 = 0.15 M

V1 * M1 = V2 * M2

V1 * 0.15 M = 0.025 L * 0.045 M

V1 = 7.5 ml

so 7.5 ml of stock solution we have prepared should be added with (25- 7.5) = 17.5 ml of water to dilute it to the required dilution of 25 ml of 0.045 M solution

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