To calculate the concentration of the new solution, we can use the formula:
\(M_1V_1 = M_2V_2\)
Given:
\(M_1 = 0.20 \, M\) (initial concentration)
\(V_1 = 6.0 \, mL\) (initial volume)
\(V_2 = 6.0 \, mL + 4.5 \, mL = 10.5 \, mL\) (final volume)
Rearranging the formula gives:
\(M_2 = \dfrac{M_1V_1}{V_2}\)
Plugging in the given values:
\(M_2 = \dfrac{0.20 \, M \times 6.0 \, mL}{10.5 \, mL}\)
\(M_2 = \dfrac{1.20}{10.5}\)
\(M_2 \approx 0.114 \, M\)
Therefore, the new concentration of the NaCl solution is approximately 0.114 M.
If 6.0 mL of 0.20 M NaCl is diluted by addition of 4.5 mL water, calculate the M concentration of the new solution.
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