Question

Dilution- 1 a) A 4 N NaOH solution was mixed with an equal volume of a 10 N NaOH. What is the concentration of NaOH in the mixture?

b) Which contains more alcohol : a 1/6 dilution of 30% ethanol or a 0.08 dilution of 60% ethanol?

C) You have a 0.3ml of a vitamin solution. How many ml of a 1/20 dilution will this make? Could you use the resulting 1/20 dilution to make a 1/100 dilution ?

D) how could you make a 4.5/5 dilution ?

PLEASE ShOw steps !! Thank you!

Answer 1

Ans. #a. Let the volume of each solution be X liters.

Total volume of mixed solution = X liters + X liters = 2X liters

Gram-equivalent of NaOH in solution = Volume x Normality = X liters x 4N = 4X gm-eq.

Gram-equivalent of NaOH in solution = X liters x 10 N = 10X gm-eq.

Total gram-equivalent on mixture = 4X gm-eq. + 10X gm-eq. = 14X gm-eq.

Now,

Normality of mixture = Number of gram-equivalents / Volume of solution in liters

                                                = 14X gm-eq. / 2X liters

‘                                               = 7 gm-eq. / liters

                                                = 7 N

#B. Final concertation of solution 1 = (1/6) of 30% = 5 %

Final concertation of solution 2 = (0.08) of 60% = 4.8 %

Therefore, total ethanol concertation in solution 1 is 5% whereas that in solution 2 is 4.8 %.

Therefore, solution 1 contains more ethanol (5%) than solution 2 (4.8%).

#C. Let the volume to be made = X

Dilution factor = Volume of original solution taken / Final volume made upto

            Or, 1/20 = 0.3 mL / X

            Or, X = 0.3 mL / (1 / 20) = 6.0 mL

Therefore, 0.3 mL can be diluted to a maximum volume of 6.0 mL to yield a (1/20) dilution.

#. Yes. A solution of higher concertation (= less dilution factor, 1/20) can be used to prepare a diluted solution (larger dilution factor, 1/100) as described above.

#D. A 4.5 / 5 dilution means there is 4.5 mL of original sample in total volume of 5.0 mL.

So, there is 4.5 mL original solution.

            5.0 mL total volume.

Now,

Amount of water (diluent) required to be added = Total vol. – Vol. of original soln.

                                                            = 5.0 mL – 4.5 mL

                                                            = 0.5 mL

Preparation: Take 4.5 mL of original solution in a clean test tube. Add 0.5 mL diluent to it to make the final solution to5.0 mL.

The resultant solution has dilution factor of 4.5 / 5.

Such solution can also be prepared by taking the respective liquids in same ratio; like 9.5 mL original solution and 1.0 mL diluent.