To what volume should you dilute 126 ml of an 7.90 M CuCl2 solution so that 49.5 ml of the diluted solution contains 5.9 g CuCl2?
To find the volume to which the 126 ml of the 7.90 M CuCl2 solution should be diluted, we can use the dilution formula:
\(C_1V_1 = C_2V_2\)
Where:
\(C_1\) = initial concentration of CuCl2 solution
\(V_1\) = initial volume of CuCl2 solution
\(C_2\) = final concentration of CuCl2 solution
\(V_2\) = final volume of CuCl2 solution
We know the following values:
\(C_1\) = 7.90 M
\(V_1\) = 126 ml
\(C_2\) = ?
\(V_2\) = 49.5 ml
First, let's calculate the amount of CuCl2 present in the 5.9 g of CuCl2, using its molar mass:
\(1 \text{ mole of CuCl2} = 63.55 \text{ g}\) (atomic mass of Cu = 63.55 g/mol)
\(5.9 \text{ g of CuCl2} = \frac{5.9}{63.55} \text{ moles of CuCl2}\)
Now, we can use the molar ratio of CuCl2 to find the number of moles in the diluted solution.
Since CuCl2 has a 1:1 stoichiometric ratio, the number of moles of CuCl2 in the diluted solution is also \(\frac{5.9}{63.55}\) moles.
Next, we can substitute the known values into the dilution formula:
\(C_1V_1 = C_2V2\)
\(7.90 \times 126 = C_2 \times 49.5\)
Rearranging the equation to solve for \(C_2\):
\(C_2 = \frac{7.90 \times 126}{49.5}\)
Now, let's calculate the final concentration (\(C_2\)):
\(C_2 = 20.08 \text{ M}\)
To find the final volume of the CuCl2 solution, we can rearrange the equation to solve for \(V_2\):
\(C_1V_1 = C_2V_2\)
\(V_2 = \frac{C_1V_1}{C_2}\)
\(V_2 = \frac{7.90 \times 126}{20.08}\)
Calculating the value of \(V_2\):
\(V_2 = 49.7 \text{ ml}\)
Thus, you would dilute the 126 ml of the 7.90 M CuCl2 solution to a volume of 49.7 ml.