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What is the concentration of NaCl, in moldm−3, when 10.0 cm3 of 0.200 moldm−3 NaCl solution is added to 30.0 cm3 of 0.600 moldm−3 NaCl solution?

Question

What is the concentration of NaCl, in moldm−3, when 10.0 cm³ of 0.200 moldm−3 NaCl solution is added to 30.0 cm³ of 0.600 moldm−3 NaCl solution?

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Solution

To solve this problem, we need to first find the total amount of moles of NaCl in each solution, and then find the total volume of the solution.

Step 1: Find the total moles of NaCl in each solution The moles of NaCl in each solution can be found by multiplying the volume of the solution by its concentration.

For the 10.0 cm3 of 0.200 moldm−3 NaCl solution: Moles of NaCl = Volume (in dm3) x Concentration (in moldm−3) = 10.0 cm3 x (1 dm3/1000 cm3) x 0.200 moldm−3 = 0.002 mol

For the 30.0 cm3 of 0.600 moldm−3 NaCl solution: Moles of NaCl = Volume (in dm3) x Concentration (in moldm−3) = 30.0 cm3 x (1 dm3/1000 cm3) x 0.600 moldm−3 = 0.018 mol

Step 2: Find the total moles of NaCl Total moles of NaCl = Moles in first solution + Moles in second solution = 0.002 mol + 0.018 mol = 0.020 mol

Step 3: Find the total volume of the solution Total volume = Volume of first solution + Volume of second solution = 10.0 cm3 + 30.0 cm3 = 40.0 cm3 Convert this to dm3 by dividing by 1000: = 40.0 cm3 x (1 dm3/1000 cm3) = 0.040 dm3

Step 4: Find the concentration of NaCl in the combined solution Concentration = Total moles / Total volume = 0.020 mol / 0.040 dm3 = 0.500 moldm−3

So, the concentration of NaCl in the combined solution is 0.500 moldm−3.

This problem has been solved

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