Chemistry question

Question

A solution is made by adding 1.23 mol of KCl to 1000.0 g of water. Assume that the

density of water (and the resulting solution) is 1.00 g/mL.

a. Calculate the formula weight of KCl.

b. Calculate the mass of KCl in grams.

c. Calculate the percent by mass of KCl in this solution.

d. Calculate the molarity of the solution.

Answer 1

a). Formula weight of KCl = 39.1 g/mol

b). Mass of KCl = 1.23 moles x 39.1 g/mol = 48.09 g = 48.1 g

c). % by mass of KCl: The total mass of solution = 48.1 g (KCl) + 1000.0 g water = 1048.1 g. Percent by mass KCl = mass KCl/total mass (x100%) = 48.1/1048.1 (x100%) = 4.59%

d). Molarity of KCl = moles KCl/liter of solution. If we assume that the KCl did not increase the volume (which is probably not a valid assumption), then you have 1.23 moles KCl in 1000 g and since 1 g = 1 ml, that's the same as 1.23 moles KCl/1000 ml = 1.23 mole/L = 1.23 M. If the question meant to ask for the molality (not molarity), then it would be moles KCl/Kg water = 1.23 moles/1 kg = 1.23 molal.