Question

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent? A.) 126 mL of 0.600 M KCl B.) 115 g of 0.440 m KCl C.)115 g of 5.0 % KCl solution by mass D.)How much solvent would you take to prepare the solution in part B? E.) How much solvent would you take to prepare the solution in part C? *please help so lost. Thanks!

Answer 1

A). 126ml 0.600 M KCl

126 ml KCl = 0.126 L KCl

0.126 L X 0.600 mol/L = 0.0756 mol KCl X 74.55 g/mol = 5.71 grams KCl

B). 0.440 m KCl means that there is 0.440 mol KCl dissolved in 1.00 kg water.

Without knowing the density of the solution, we cannot really calculate the mass of KCl in 129 g of the solution. However, if we assume that the density of the solution is 1 g/mL, then:

115 grams KCl = 0.115 kg KCl

0.115 kg solution X 0.440 mol KCl / 1 kg solvent = 0.0506 mol KCl.

molar mass of KCl = 74.55 g/mol

therefore, mass of dry solute = 0.0506 × 74.55 = 3.77 grams

C). 5% (m/m) means that 100 g of the solution will contain 5 g KCl.
Therefore, 115 g solution X (5 gram KCl/100 gram solution) = 5.75 g KCl

D). To prepare solution in part B:
0.0506 mol KCl / 0.440 mol KCl/1 kg H2O = 0.115 kg water = 115 mL water

E).To prepare the solution in part C:
115 g of solution - 5.75g KCl = 109.25 g H₂O