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.
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.
To calculate the concentration of the new solution, we need to consider the moles of solute before and after the dilution.
Step 1: Calculate the moles of NaCl before dilution:
Moles = Concentration x Volume
Moles = 0.20 M x 6.0 mL
Moles = 1.2 mmol
Step 2: Calculate the total volume of the new solution:
Total Volume = Initial Volume + Volume of Water added
Total Volume = 6.0 mL + 4.5 mL
Total Volume = 10.5 mL
Step 3: Calculate the concentration of the new solution:
Concentration = Moles / Total Volume
Concentration = 1.2 mmol / 10.5 mL
To convert this concentration to Molarity, we divide by 1000 (to convert mL to L) and then divide by the molecular weight of NaCl (58.44 g/mol) for the molar conversion.
Concentration (Molarity) = (1.2 mmol / 10.5 mL) / (1000 mL/L) / (58.44 g/mol)
Concentration (Molarity) ≈ 0.002 M
Therefore, the Molarity of the new solution is approximately 0.002 M.