Question

(a) How many grams of CaCl2 are needed to make 798.0 g of a solution that is 32.5% (m/m) calcium chloride in water? Note that mass is not technically the same thing as weight, but (m/m) has the same meaning as (w/w).

-How many grams of water are needed to make this solution?

(b) What is the volume percent % (v/v) of an alcohol solution made by dissolving 147 mL of isopropyl alcohol in 731 mL of water? (Assume that volumes are additive.)

(c) The mass of solute per 100 mL of solution is abbreviated as % (m/v). (The abbreviation % (w/v) is also common.) How many grams of sucrose are needed to make 915 mL of a 38.0% (w/v) sucrose solution?

Answer 1

(a)32.5% (m/m) calcium chloride in water means 32.5% of the total mass of the solution is CaCl₂

So mass of CaCl₂ required is = 32.5 % x total mass of the solution
                                            = (32.5 / 100 ) x 798.0 g
                                          = 259.35 g
                                              = 259 g

So mass of CaCl₂ needed is 259 g

mass water in the solution is = total mass of the solution - mass of CaCl₂
                                             = 798.0 - 259   g
                                             = 539 g

(b) Given volume of isopropyl alcohol is , V = 147 mL

               Volume of water , V' = 731 mL

Total volume of the solution , V'' = V + V'

                                                   = 147 + 731

                                                  = 878 mL

So volume percent % (v/v) of an alcohol is = ( volume of alcohol / volume of solution ) x 100

                                                                   = ( 147 / 878 ) x 100

                                                                  = 16.7 %

Therefore the volume percent % (v/v) of an alcohol is 16.7 %

(c) 38.0% (w/v) sucrose solution means 38.0 g of sucrose present in 100 mL of solution

                                                               Z g of sucrose present in 915 mL of solution

                                                                           Z = ( 38.0 / 100 ) x 915

                                                                             = 347.7 g

So mass of sucrose needed is 347.7 g