Chemistry/ stoichipmetry

Question

3.9g KCl in 75.0mL of 0.240M CaCl2 solution. Assume that the volumes are addictive. What is MK+,MCa+, and MCl-?

Answer 1

There may be information missing from this problem. It states "volumes are additive", but there is only one liquid solution (one volume) specified in the problem. The KCl is a solid. Nonetheless here is the solution to the problem as written:

Determine the amount of KCl and CaCl₂ in mol:

for KCl use the molar mass to convert from mass in g to mol; from the periodic table 1 mol K = 39.098 g and 1 mol Cl = 35.45 g; the molar mass (1 mol) KCl = 39.098 g + 35.45 g = 74.55 g KCl

Given 3.9 g KCl use the molar mass as the conversion factor in dimensional analysis to calculate mol KCl:

3.9 g KCl x ^{1 mol KCl} / _{74.55 g KCl} = 0.052 mol KCl

There is 1 mol K⁺ in 1 mol KCl, so 0.052 mol KCl x ^{1 mol K+} / _{1 mol KCl} = 0.052 mol K⁺

molarity (M) = ^{mol solute} / _{volume solution (L)} and the volume of the solution is 75.0 mL x ^{1 L} / _{1000 mL} = 0.075 L

Concentration (M) = ^{0.052 mol K+} / _{0.075 L} = 0.69 M K⁺

In 1 mol CaCl₂ there is 1 mol Ca⁺² and 2 mol Cl^–

^{0.240 mol CaCl2} /_{1 L solution} x ^{1 mol Ca+2} /_{1 mol CaCl2} = ^{0.240 mol Ca+2}/ _{L solution} = 0.24 M Ca⁺²

^{0.240 mol CaCl2}/_{1L solution} x ^{2 mol Cl–}/_{1 mol CaCl2} = ^{0.480 mol Cl–}/_{L solution} = 0.48 M Cl^–