A solution is made by adding 5.0 g Na2CO3 * 10H2O to enough water to make a 500 mL solution. what is the molarity.
I have: M=mol solute/liter solution = (5/286)/.5 = .035M
286 is what i got for the molar mass of Na2CO3 * 10H2O
Is that correct?? I think it is, but the next problem refers back to that problem and makes me think i did something wrong:
How much of this 500 mL solution should be used and diluted to make 250 mL of a .035 M solution
if the concentration is already .035 shouldnt it just be 250 mL????????
It looks ok to me.
Your calculation for the molarity of the solution seems correct. The molar mass of Na2CO3 * 10H2O (sodium carbonate decahydrate) is indeed 286 g/mol, so to find the moles of solute, we divide the mass of Na2CO3 * 10H2O (5.0 g) by its molar mass (286 g/mol). Then, we divide the moles of solute by the volume of the solution in liters (500 mL = 0.5 L) to get the molarity, which in this case is 0.035 M.
Now, moving on to your second question, you are correct that if you want to make a solution with the same concentration as the initial solution (0.035 M), and you only need a smaller volume (250 mL), you can simply take 250 mL of the original solution and dilute it further. The concentration will remain the same because the number of moles of solute will remain constant. Therefore, in this case, you would indeed use only 250 mL of the 0.035 M solution.
It seems like there might be an error in the formulation of the second problem you mentioned. If the concentration is already 0.035 M and you are asked to make a solution of the same concentration but with a smaller volume, you can simply use that smaller volume without needing to dilute it further.