Write a detailed procedure for preparing 50.0 mL of 1.20 M HCl. For this procedu

Question

Write a detailed procedure for preparing 50.0 mL of 1.20 M HCl. For this procedure you have only the following supplies available (plus any safety gear, equipment, and supplies required): a 10 mL graduated cylinder, a 50.0 mL volumetric flask, many 100 mL beakers, DI water, and 12.0 M HCI. 7. Write a detailed procedure for preparing 50.0 mL of 0.0120 M HCI. For this procedure you have only the following supplies available (plus any safety gear, equipment, and supplies required): a 10 mL graduated cylinder, a 50.0 mL volumetric flask, many 100 mL beakers. DI water, and 12.0 M HCI.

Explanation / Answer

6)

Step 1: Calculation

molarity of stock HCl solution , M1 = 12.0M

molariy of HCl solution to be prepared, M2 = 1.20M

Volume of the HCl solution to be prepared, V2 = 50mL

Volume of the stock HCl solution to be taken , V1 = ?

M1 V1 = M2 V2

V1 = M2 V2 / M1

V1 = (1.20 x 50) / 12

V1 = 5 mL

So, We have to take 5mL of 12.0 M HCl solution to make 50mL of 1.20 M HCl solution.

Step 2 : Procedure

7.

Step 1: Calculation

molarity of stock HCl solution , M1 = 12.0M

molariy of HCl solution to be prepared, M2 = 0.0120 M

Volume of the HCl solution to be prepared, V2 = 50mL

Volume of the stock HCl solution to be taken , V1 = ?

M1 V1 = M2 V2

V1 = M2 V2 / M1

V1 = (0.0120 x 50) / 12

V1 = 0.05 mL

So, We have to take 0.05mL of 12.0 M HCl solution to make 50mL of 0.0120 M HCl solution.

Since, we can't use 10mL graduated cylinder to measure 0.05mL , therefore, we will first make 1.20 M HCl solution then we will dilute it to make 0.0120M HCl solution.

To make 50mL of 1.20 M HCl solution from 12.0 M HCl solution, calculation is same as in the question 6. we will take 5mL of 12.0 M HCl solution.

Calculation to make 50 mL of 0.012 m HCl solution using 1.20 M HCl solution:

Step 1: Calculation

molarity of stock HCl solution , M1 = 1.20 M

molariy of HCl solution to be prepared, M2 = 0.0120 M

Volume of the HCl solution to be prepared, V2 = 50mL

Volume of the stock HCl solution to be taken , V1 = ?

M1 V1 = M2 V2

V1 = M2 V2 / M1

V1 = (0.0120 x 50) / 1.2

V1 = 0.5 mL

So, We have to take 0.5mL of 1.20 M HCl solution to make 50mL of 0.0120 M HCl solution.

Step 2 : Procedure

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