How many grams os Al2(SO4)3 are needed to prepare 250.0 mL of a 0.400 M Al2(SO4)3 solution...ANSWER IS: 34.21g
QUESTIONS IS: What would the concentration be if we diluted this soultion to 1000.0 mL?
Thanks you guys soooo much...I really appreciate it!
2006-12-11 12:06:34 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
i dont know
2006-12-11 12:08:11 · answer #1 · answered by mack smith 2 · 0⤊ 0⤋
copy from http://answers.yahoo.com/question/index?qid=20061128104732AAZmQlX
"You want to have 0.250 moles of Al2(SO4)3 in each liter of solution. You're preparing 0.4 L of solution, so you need (0.4 * 0.250) = 0.1 moles of Al(SO4)3.
Now the molar mass of each element is listed on most periodic tables. That's how much one mole of that atom weighs. A molecule weighs as much as all the atoms in it put together, so a mole of Al2(SO4)3 weighs as much as two moles of Al, 3 moles of S and 12 moles of O. That's 2(26.982) + 3(32.06) + 12(15.999), according to my periodic table. So if you add it all up, a mole of Al2(SO4)3 weighs 342.132 grams. You only need a tenth of a mole, so you use 34.2132 grams."
2006-12-11 12:17:42 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Based on the formula mv= mv where m is the molarity of the solution and v is the volume :
0.400 mol/L X .250L = m X 1L
If you solve for the missing m the molarity of the new solution would be 0.1 mol/L
2006-12-11 12:19:41 · answer #3 · answered by Peter W 2 · 0⤊ 0⤋
c=m/v, m=mass
m remains constant.
so, c1v1=c2v2
so, after diluting the solution 4 times its original volume, the final conc would be c2=0.4/4=0.1M.
2006-12-11 12:21:16 · answer #4 · answered by DC 4 · 0⤊ 0⤋