What volume in L of a 0.320 M Mg(NO₃)₂ solution contains 45.0 g of Mg(NO₃)₂ ?

Question

While doing this problem I get stuck at a certain point. If I am wrong from the start please correct me. I know the molar mass of  Mg(No3)2 = 148.3 g/mol. but from there I get stuck. Any help is appreciated.

Answer 1

Hello Daniel,

You are correct by needing the molar mass of the compound, however there are a few more details to lead you on the proper track. This question is asking you to calculate the volume of a solution given the amount of substance and the molarity. We know from the equation for molarity that (molarity = number of moles/ liter). You are given the molarity, which is 0.320 M, and the grams of the compound which is 45.0g. Given that moles= grams of compound/ molar mass, we can calculate that the number of moles of Mg(NO3)2 is 45.0g/148.3g/mol which equates 0.303 moles. From our original equation for molarity we can now substitute the known molarity and number of moles to solve for the volume. 0.320M= 0.303 moles/ (x)Liters. The number of liters of solution is 1.05.

I hope this helps.