Question

1.Less quantitative dilutions can also be done without a volumetric flask, The precision of concentration is dependent on how precise the volumes are transferred. For example a dilute solution of HCl can be made using a precise amount of 5.0 M HCl and distilled water. In this case the total volume of dilute HCl (V2) is the total volume of 5.0 M HCl plus the volume of water. Determine the volume of water in mL needed to prepare 5.0mL (V2) of a 3.1 M (M2) HCl solution using 6.2 M (M1) HCl (M1).

2.Less quantitative dilutions can also be done without a volumetric flask, The precision of concentration is dependent on how precise the volumes are transferred. For example a dilute solution of HCl can be made using a precise amount of 5.0 M HCl and distilled water. In this case the total volume of dilute HCl (V2) is the total volume of 5.0 M HCl plus the volume of water. Determine the volume (V1) of 6.8 M (M1) HCl in mL needed to prepare 5.0 mL (V2) of 0.6 M (M2) HCl solution.

3.What is the molarity of HCl in a solution if it is prepared when 1.74 mL of 3.06 M HCl is mixed with 1.34 mL of water?

Answer 1

Solution :-

Q1)

Final volume V2 = 5.0 ml

Initial molarity M1= 6.2 M

Final molarity M2= 3.1 M

Volume of Water= ?

Lets first calculate the volume of the stock solution we need to make the desired dilute solution

M1V1=M2V2

V1=M2V2/M1

V1=3.1 M * 5.0 mL / 6.2 M

V1 = 2.5 mL

Now using the final volume and initial volume we can find the volume of water

Volume of water = final volume – initial stock volume

                               = 5.0 mL – 2.5 mL

                                = 2.5 mL

Therefore we need 2.5 mL water

Q2)

Initial molarity M1 = 6.8 M

Initial volume V1 = ?

Final volume V2 = 5.0 mL

Final molarity M2 = 0.6 M

M1V1=M2V2

V1 = M2V2/M1

      = 0.6 M * 5.0 mL / 6.8 M

      = 0.44 mL

Therefore we need to use 0.44 mL of stock solution to make the dilute solution.

Q3)

Initial volume V1 = 1.74 mL

Initial molarity M1 = 3.06 M

Final molarity M2= ?

Final volume = 1.74 mL + 1.34 mL = 3.08 mL

M1V1=M2V2

M1V1/V2 = M2

3.06 M * 1.74 mL / 3.08 mL = M2

1.73 M = M2

Therefore the final concentration of the HCl is 1.73 M