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.
Dilution- 1 a) A 4 N NaOH solution was mixed with an equal volume of a...
1. What volume of 0.10 M NaOH is needed to neutralize 100 mL of 0.050 M HCl? 2a. When 100 mL of 0.01 M NaOH solution is mixed with 100 mL of 0.01 M of HNO3 solution, what is the concentration of H+ and the pH in the resulting mixture? 2b. When 100 mL of 0.01 M NaOH solution is mixed with 100 mL of 0.02 M of HCl solution, what is the concentration of H+ and the pH in...
A 10 mL volume of a 5 mM lactic acid solution is mixed with 5 mL of water. What is the concentration of the resulting mixture in
Question 1 20 pts Please prepare 20 mL of 0.2 M NaOH, given a stock solution of 0.6 M NaOH. State the precise amount of the 0.6 M NaOH used and the amount of diluent (H2O) added Question 2 20 pts Please prepare 500 mL of a 1/100 dilution of serum from a tube with a 1/10O dilution of serum. Please state the volume of the 1/10 dilution of serum used and the volume of diluents (H2O) used 20 pts...
Lab 5: Additional Dilution Problems You have a stock solution that contains 24 mg/ml of protein. You want to make a set of standards with the following concentrations: 12 mg/ml, 6 mg/ml, 3 mg/ml and 1.5 mg/ml You need a minimum of 500 ul of each standard after all dilutions are completed. Using the technique of serial dilution, explain how you would make your standards by filling in the chart below. Std 4 Std 1 12 Std 2 6 Std...
6. In a calorimeter, HCl is mixed with NaOH at 20 °C and the volume of the resulting solution is 180 mL. When the reaction is complete the temperature of the solution is 26 °C. Determine the heat for the reaction assuming the specific heat of water is 4.18J/gºC, and the density of the solution is 1.2 g/mL. a. 5417) b. 4514) c. -5417) d. -4514)
Serial Dilution Questions 1, Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution of dye that has a concentration of 350.0 mg/L and do the following set of serial dilutions on it: 5.00 mL of stock diluted to a total volume (TV) of 500.00 mL (Dilution 1); 2,00 mL of Dilution 1 is then diluted to a TV of 250.00 mL (Dilution 2); finally, 1.00 mL of Dilution 2 is then diluted to a total volume of...
Electrolytes, Concentration and Dilution of Solutions n this part of the experiment, you will make a dilution of a stock solution of known concentration and calculate the concentration of the dilute solution. DILUTION OF A STOCK SOLUTION PART C: Measurements and Observations Procedure Check out a 100-mL volumetric flask from your instructor. Fill test tube about 2/3 concentration of the active ingredient from its label including units. Record 1. orlt O " concentration of active ingredient. full of "Magic 2....
Can you make up 20 ml of a 1:200 dilution from 2 ml of a 1:40 stock solution? (a) If so, how? If not why not? (b) If not what is the largest volume of a 1:200 you could make?
1 of 4) Watch the video that shows using the dilution method to make a solution of known concentration. Note that while the video shows the use of a beaker, in the lab we typically use a volumetric flask to make the diluted solution. The stock solution of KMnOhas a concentration of 2.6 x 104m. The pipette has a volume of 10.0 mL. What is the amount of KMnO4 delivered to the solution, in moles? mol Submit
1. A certain experiment requires 624.0 mL of 0.00445 M NaOH. i). Calculate the number of grams of NaOH required to make this solution. ii). Calculate the mass percentage of a 0.00445 M NaOH solution (the density of solution is 1.52 g/mL). iii). Calculate the pOH and pH of this 0.00445 M NaOH solution. iv). What volume of 0.175 M NaOH solution is required to make 624.0 mL of 0.00445 M NaOH solution? v). Calculate the pOH and pH of...