Question

o Calculate the following sample concentrations based upon given dilutions. Final (Diluted) Concentration 1st Dilution 2nd Dilution Concentration Original 1:5 1:4 1:15 1:3 1:3 3 (none) 1:2 1:5 (none) 1:5 167 mg/dL b. 139 mg/dL 15 mg/dL d. 557 mg/dL 23 mg/dL Page 2 of 3

Answer 1

Answer 1. In this case the original concentration is 167 mg/dL and after its 1st dilution of 1:5, the final concentration would be; 167x1/5 or 167/5= 33.4.

Thus the final concentration would be 33.4 mg/dL

Answer 2. In this case the original concentration is 139 mg/dL and its 1st dilution is 1:4 i.e 139x 1/4, which means 139/4 and is 34.75 mg/dL. Now the second dilution is of 1:2 which goes as 34.75x 1/2, or 34.75/2 and comes to 17.375 mg/dL.

Thus, here the final concentration would be 17.375mg/dL.

Answer 3. Here the final concentration is already given but the original is to be find out. So starting from the final concentration we will calculate as it as:

Ax1/5=15; A=15x5; A=75. Proceeding further, Bx1/15=75; or B= 75x15 that will become, B= 1125.

So here the original concentration would be of 1125 mg/dL.

Answer 4. In this case, the original concentration is of 557mg/dL and is done with 1:3 dilution that means, 557x1/3 or 557/3 i.e 185.66

So, the final concentration here would be 185.66 mg/dL.

Answer 5. Here again the final concentration is provided but not the original concentration. We will again back calculate it as, Ax1/5=23, or A= 23x5, and A= 115.

Now Bx1/3=115, or B= 115x3, and B = 345.

So, here the original concentration would be 345mg/dL