Question

2. Determine the number of grams of KCl needed to make the following:

A. 2.50 liters of 2.00 M K solution

B. 133 g of a 0.900% by mass KCl

C. 133 ml of a 0.900 M KCl

D. 0.900 m KCl in 500 grams of water

E. 6.00 moles total of a 0.750 mole fraction of KCl

Answer 1

A. 2.50 liters of 2.00 M K solution

no of moles KCl = molarity * volume in L

                            = 2*2.5

                            = 4.5 moles

mass of KCl         = no of moles * gram molar mass

                             = 4.5*74.5 = 335.25g

B. 133 g of a 0.900% by mass KCl

    0.9% KCl solution means 0.9g of KCl present in 100g of solution

                                            100g of solution contains 0.9g of KCl

                                             133g of solution contains = 0.9*133/100   = 1.197g of KCl

C. 133 ml of a 0.900 M KCl

no of moles of KCl   = molarity * volume in L

                                 = 0.9*0.133   = 0.1197 moles

mass of KCl             = no of moles * gram molar mass

                                = 0.1197*74.5   = 8.9g

D. 0.900 m KCl in 500 grams of water

molality   = W*1000/G.M.Wt* weight of solvent in g

0.9           = W*1000/74.5*500

W            = 0.9*74.5*500/1000   = 33.525g

E. 6.00 moles total of a 0.750 mole fraction of KCl

    mole fraction = no of moles of one component/Total no of moles

                          = 0.75/6   = 0.125