What is the molarity of the following solutions?

Question

(a) 0.78 g MgCl₂ in 700 mL of solution 

(b) 9.47 g Na₂S in 300 mL solution

(c) 1.73 g CaF₂ in 315 mL of solution

Answer 1

a) grams of MgCl2 divided by its formula weight gives you the moles present for that compound. To find the molarity, you will divide the moles of MgCl2 by the volume in Liters which means you have to convert those 700 mL to Liters.

moles of MgCl2 = 0.78g / 95.211 g/mol = 0.00819233 moles

Molarity of MgCl2 = 0.00819233 moles / 0.7 L = 0.01170333 M

Since the least amount of significant figure is one then the answer becomes 0.01 M of MgCl2

b) and c) you would do the same procedures as a) however with the compound mentioned in that part of the exercise