If we want to prepare 125 mL of a 1.245 M aqueous KCl solution from a 7.134 M aqueous KCl solution, how many mL of the 7.134 M solution (rounded with correct number of significant figures) will we need to put in the new solution?
The dilution formula is
mL1 x M1 = mL2 x M2
mL1 x 7.134M = 125 x 1.245
Solve for mL1 = ?
To find out how many mL of the 7.134 M KCl solution we need to put into the new 1.245 M KCl solution, we can use the dilution formula:
\(c_1V_1 = c_2V_2\)
Where:
\(c_1\) is the concentration of the initial solution (7.134 M)
\(V_1\) is the volume of the initial solution (unknown)
\(c_2\) is the concentration of the final solution (1.245 M)
\(V_2\) is the volume of the final solution (125 mL)
Rearranging the formula to solve for \(V_1\), we get:
\(V_1 = \frac{{c_2V_2}}{{c_1}}\)
Plugging in the values, we have:
\(V_1 = \frac{{1.245 \, \text{M} \times 125 \, \text{mL}}}{{7.134 \, \text{M}}}\)
Calculating this expression, we find:
\(V_1 = 21.68 \, \text{mL}\)
Therefore, we need to put approximately 21.68 mL of the 7.134 M KCl solution into the new solution. Since we are rounding to the correct number of significant figures, the answer is 22 mL.