Suppose you diluted a concentrated salt solution in the foll

Suppose you diluted a concentrated salt solution in the following manner - take 2 ml of the salt solution and add to 198 ml of water, mix it. then you take 1 ml of this mix, add it to 9 ml of water and mix it. Then you take 2 ml of this and added it to 8 ml of water and mix it. You find that the final dilution of salt solution contains 20 ng NaCl /ml. What was the concentration of NaCl in your original solution?

Solution

Ans. Let the original salt concentration be X g/ mL.

#1. 2 ml of original salt solution is mixed with 298 mL distilled water.

            Amount of salt in 2 mL = concentration x volume of solution taken

                                                = X g/Ml x 2 mL = 2X g

            Total volume of solution made upto = 2 mL (salt solution) + 198 mL (distilled water)

                                                = 200 mL

Final concentration of resultant solution #1 = Amount of salt / final volume

                                                = 2X g/ 200 ml

                                                = 0.001 X g/mL

#2. Amount of salt in 1.0 mL of solution #1 = Concentration of solution #1 / volume of solution taken

                                                = (0.001 X g/mL) x 1.0 mL

                                                = 0.001 X g

Final volume made upto = 1.0 mL (#1) + 9 mL (water) = 10 mL

Concentration of resultant solution #2 = 0.001 X g / 10.0 mL = 0.0001 X g /mL

#3. Amount of salt in 2.0 mL of solution #2 = Concentration of solution #2 / volume of solution taken

                                                = (0.0001 X g/mL) x 2.0 mL

                                                = 0.0002 X g

Final volume made upto = 2.0 mL (#2) + 8 mL (water) = 10 mL

Concentration of resultant solution #3 = 0.0002 X g / 10.0 mL = 0.00002 X g /mL

Now,

            Given, concentration of #3 = 20 ng/ mL

Also, we have calculated, concentration of #3 = 0.00002 X g /mL

Thus, 0.00002 X g /mL = 20 ng/ mL = 20 x 10^-9 g/mL                              ; [1 ng = 10^-9 g]

                        Or, X = 20 x 10^-9 g/mL / (0.00002 X g /mL)

                        Hence, X = 1 x 10^-3

Therefore,

Concentration of original salt solution = X g/mL

                                                            = 1 x 10^-3 g /mL = 0.001 g/mL

                                                            = 1.000 mg/ mL                         ; [1 g = 1000 mg]