moles CuCl2 = 5.9 g / molar mass CuCl2
molarity = moles CuCl2 / .0500 L
diluted volume = (8.00 M / molarity) * 125 mL
To what volume should you dilute 125mL of an 8.00M CuCl2 solution so that 50.0mL so
that 50.0 mL of the diluted solution contains 5.9 g CuCl2?
2 answers
8.0 M CuCl2 contains 8 molsCuCl2/L of solution. ALL of the following calculations are rounded so if you need more accurate number you should recalculate everything including the atomic masses and molar masses.
1 mol CuCl2 = 63.5 + 2*35.5 = approximately 134 g; therefore, there are 8.0 x 134 = about 1072 grams CuCl2 in 1 L.
moles CuCl2 in 5.9 g = 5.9/134 = 0.044. That in 50 mL is (0.044/0.050) = 0.88 M.
The dilution formula is mLa x Ma = mLb x Mb.
125 mL x 8.0 M = mLb x 0.88 M
mL b = 125 x 8.0/0.88 = 1,136 mL.; therefore, take 125 mL of the 8.0 M CuCl2 and dilute it to 1,136 mL.
That should do it. Check it.
8.0 M CuCl2 x 125 mL/1,136 mL = 0.88 M or 0.88 mols/L.
0.88 mols/L x 0.050 L = 0.044 moles in the 50 mL.
Then 0.044 x 134 g/mol = 5.897 = essentially 5.9. It will be 5.9 if you clean up the math and the rounding etc. Post your work if you get stuck.
1 mol CuCl2 = 63.5 + 2*35.5 = approximately 134 g; therefore, there are 8.0 x 134 = about 1072 grams CuCl2 in 1 L.
moles CuCl2 in 5.9 g = 5.9/134 = 0.044. That in 50 mL is (0.044/0.050) = 0.88 M.
The dilution formula is mLa x Ma = mLb x Mb.
125 mL x 8.0 M = mLb x 0.88 M
mL b = 125 x 8.0/0.88 = 1,136 mL.; therefore, take 125 mL of the 8.0 M CuCl2 and dilute it to 1,136 mL.
That should do it. Check it.
8.0 M CuCl2 x 125 mL/1,136 mL = 0.88 M or 0.88 mols/L.
0.88 mols/L x 0.050 L = 0.044 moles in the 50 mL.
Then 0.044 x 134 g/mol = 5.897 = essentially 5.9. It will be 5.9 if you clean up the math and the rounding etc. Post your work if you get stuck.
1 answer