Dried Mik Pwders Data Sheet concentration of EDTA solution m

Dried Mik Pwders Data Sheet concentration of EDTA solution, mol Lga brand name or identification code of milk powder sample cost of 1 oz of milk powder sample, S s determination initial mass of weighing bottle and milk powder,g final mass of weighing bottle and contents after transferring sample, g/ mass of powder transferred, g final buret reading, mL initial buret reading, ml IS as sm o onml .oni volume of EDTA solution required 1M235 number of moles of Ca number of grams of Ca2 ion mass percent Ca2+ ion in sample % for titration, ml. in sample, mol in sample, g ion mean mass percent Ca2+ ion in sample, % mean cost per ounce of Ca2 ion in sample,Soz- Summary of Class Data brand of milk powderK cost per ounce of Ca? ion, $

Solution

Ans) The resultant complex that forms has a 1:1 metal is to edta ratio.

For sample 1: volume of edta used is 27.2 ml and concentration of edta is 0.02M. So the number of moles of edta used is (27.2/1000) * 0.02 = 0.544 milimoles = moles of calcium. Number of grams of calcium is 40 * 0.000544 = 0.02176 grams.

Mass percent of calcium is (0.02176/1.51)* 100 = 1.44%

For sample 2:

Volume of edta used is 23.5ml

Number of moles of edta used is (23.5/1000) * 0.02= 0.47 milimoles= number of moles of calcium.

Mass of calcium present is 40* 0.00047 = 0.0188 grams

Mass percent of calcium in the sample is (0.0188/1.45) *100= 1.29%

Mean mass percent is 1.365%

The data for cost per ounce of sample is vaguely visible. I can only sort out the logic to calculate it.

1 ounce of sample will have 0.013 ounce of calcium. Then we should calculate the amount of sample corresponding to one ounce calcium and then multiply with the cost of sample per ounce to get the required answer.