Question

If 3.0 liters of 0.50 m nacl (aq) is mixed with 9.0 liters of 0.2777 m nacl (aq), what is the final concentration of the resulting solution?

Answer 1

The final concentration of NaCl after mixing 3.0 liters of 0.50 M NaCl and 9.0 liters of 0.2777 M NaCl is 0.333275 M.

Step-by-step explanation:

To calculate the final concentration of a NaCl solution after mixing two solutions with different volumes and concentrations, you need to use the formula for dilution: M1V1 + M2V2 = MfVf, where M stands for molarity, V stands for volume, and the subscript f refers to the final state.

First, we calculate the moles of NaCl in each solution before mixing:

Moles in 3.0 L of 0.50 M NaCl = 3.0 L × 0.50 M = 1.5 moles

Moles in 9.0 L of 0.2777 M NaCl = 9.0 L × 0.2777 M = 2.4993 moles

Now, we add the moles together and divide by the total volume to find the final concentration:

Total moles of NaCl = 1.5 moles + 2.4993 moles = 3.9993 moles

Total volume = 3.0 L + 9.0 L = 12.0 L
Final concentration = 3.9993 moles / 12.0 L = 0.333275 M

Answer 2

First, we shall calculate the total number of moles present in the final solution.
Number of moles in 0.50 m NaCl = molarity * volume = 0.50 * 3.0 = 1.5 moles.
Number of moles in 0.2777m NaCl = molarity * volume = 0.2777 * 9.0 = 0.24993 moles
Total number of moles = 1.5 + 0.24993 = 1.74993 moles

Second, we shall calculate the total volume of the final solution.
Total volume = 3 + 9 = 12 litres.

The molarity = total number of moles / total volume = 1.74993 / 12 = 0.1458 m