Question

1- How many milliliters of an aqueous solution of 0.122 M aluminum chloride is needed to obtain 13.9 grams of the salt?

2- In the laboratory you dissolve 15.7 g of barium sulfide in a volumetric flask and add water to a total volume of 250 . mL. What is the molarity of the solution?

3- In the laboratory you dilute 4.43 mL of a concentrated 6.00 M hydroiodic acid solution to a total volume of 50.0 mL. What is the concentration of the dilute solution?

4- In the laboratory you dilute 3.77 mL of a concentrated 3.00 M hydroiodic acid solution to a total volume of 150 mL. What is the concentration of the dilute solution?

Answer 1

1.Molar mass of AlCl3 is 133.34 g/mol

We know that Molarity , M = ( mass/molar mass) *(1000/Volume of solution in mL)

                       0.122M = (13.9g/133.34(g/mol))*(1000/V)

                           V = 854.5 mL

2.Molar mass of BaS is 169.4 g/mol

Molarity , M = ( mass/molar mass) *(1000/Volume of solution in mL)

                  = ( 15.7g / 169.4(g/mol))*(1000/250 mL)

                  = 0.371 M

3.According to law of dilution   MV = M'V'

Where M = Molarity of stock = 6.00 M

V = Volume of the stock = 4.43 mL

M' = Molarity of dilute solution = ?

V' = Volume of the dilute solution = 50.0 mL

Plug the values we get   , M' = MV /V'

                                         = ( 6.00M*4.43mL) / 50.0 mL

                                         = 0.532 M

4.According to law of dilution   MV = M'V'

Where M = Molarity of stock = 3.00 M

V = Volume of the stock = 3.77 mL

M' = Molarity of dilute solution = ?

V' = Volume of the dilute solution = 150 mL

Plug the values we get   , M' = MV /V'

                                          = ( 3.00M*3.77mL) / 150mL

                                          = 0.075 M