Calculate the molarity of the following aqueous solution of 25.0 grams of MgBr2 in 0.355 L solution?
Did I work this right?
25gMgBr2/184.113 gMgBr2
=.1357861748 mol MgBr2
then:
.1357861748 mol MgBr2/0.355 L
=.382 M
It looks ok to me but you don't need all of those intermediate numbers.
Thank you DrBob222! :)
Yes, you worked it out correctly.
To calculate the molarity of a solution, you need to know the moles of solute (in this case, MgBr2) and the volume of the solution.
First, you converted the mass of MgBr2 (25.0 grams) to moles by using the molar mass of MgBr2 (184.113 g/mol). Dividing 25.0 grams by 184.113 g/mol gives you 0.1357861748 moles of MgBr2.
Then, you divided the moles of MgBr2 by the volume of the solution (0.355 L), resulting in a molarity of 0.382 M (moles per liter).
Therefore, the molarity of the aqueous solution is 0.382 M.