## molarity

**Anonymous**

I have this homework question, I and i dont understand it. Here it is:

A student made calculations for the preparation of 1 liter of 4M barium chloride solution, but neglected to account for the 2 moles of water per forumula unit when he weighed out the solid. Was the actual molarity of the solution he prepared too high or too low?

Can anyone help me with this? and explain it to me.

Thanks

Anonymous| 22 January 2004 - 10:40pmThis one is pretty easy. First, it helps to write out the formula of the solute as a point of reference. In this case, it's BaCl2*2(H2O). The molar mass of the compound [b]without[/b] the two moles of water figured in is 208.232 g/mole. This is what the student used for his calculations.

One Liter of a 4M solution contains 4 moles of whatever the solute is. (In this case, four moles of Barium chloride). So the student measured out four times the molar mass calculated above. 4 X 208.232 = 832.930 grams. This amount of Barium chloride was put into a one Liter volumetric flask, and the flask was filled up to the line with water. This gave him what he thought was a 4 Molar solution.

As you are aware, the student did not actually make a 4 Molar solution. To figure out the actual molarity of the solution, we first have to figure out the molar mass of the Barium chloride including the two water molecules. This equals 208.232 (the MM of BaCl2) + 2(18.0153 g/mole). The total of these two values is 244.263 g/mole. THIS is what the student should have used to calculate the mass of solute needed to make a 4M solution.

Knowing the amount of BaCl2*2(H2O) the student had weighed out, 832.930 grams, we can calculate the number of moles of Barium chloride dihydrate he has by dividing the 832.930 grams by the molar mass of BaCl2*2(H2O). 832.930/244.263 = 3.40997 moles. Therefore, the solution he made in error is actually ~3.4M. The actual solution he made was too low.

By not including the water in his calculations, the mass of solute that he was using had a substantial amount of water in it. His calculations assumed that the water was BaCl2, when in fact it wasn't. When making solutions, it is imperative that you take into account parts of the solute that aren't what you want to make a solution of.