Question

. Suppose that you are preparing CaCl2 solutions.

a. What mass of CaCl2(s) would you need to add to make 100.0 mL of a 0.020 M solution?

b. If you actually added 0.228 g of CaCl2(s), what would the solution concentration be?

c. You dilute this solution 5-fold, adding 10.00 mL into a 50.00 mL volumetric flask. What is the resultant concentration?

Answer 1

molar mass of CaCl2 = 111 g/mol

a)

number of moles of CaCl2 = M*V

                           = 0.020 m * 0.100 L

                            =0.002 mol

mass of CaCl2= molar mass * number of moles

               = 111*0.002

               = 0.222 g

Answer: 0.222 g

b)

number of mol, n = mass/molar mass

                  = 0.228/111

                  =0.00205 mol

volume = 0.100 L

concentration = number of mol / volume

               = 0.00205/0.100

              =0.0205 M

Answer: 0.0205 M

c)

diluting 5 fold will give concentration =

0.0205/5

=0.00411 M

Answer: 0.00411 M