Question 1: A student determined Ca2+ ion content in a cample of nonfat dry milk

Question

Question 1: A student determined Ca2+ ion content in a cample of nonfat dry milk. For complete titration of the ion present ,a sample weighing 1.483 g required 41.33mL of 1.183e-2M EDTA solution.

a) calculate the mass of Ca2+ ion in titrated sample: 1.96e-2 g

b) a 25.6 oz box of powdered milk analyzed in (a) costs $5.46. Calculate cost of 1g of Ca2+ ion as procided by this brand of dry milk: $1.75.

Many individuals increase their Ca2+ ion intake by taking high-potency calcium tablets. A bottle of 60 tablets, each containing 1.500 g of calcium carbonate, costs $3.98.

1) Calculate the number of grams of Ca2+ ion present in ONE calcium tablet.

2) How many cups of milk, each prepared by dissolving 22.7g of the milk powder in Question 1, must a person drink to get the same amount of Ca2+ ion present in one high-potency calcium tablet?

3) Compare the cost of 1.00g of Ca2+ ion obtained from the milk powder to the cost of the same amount obtained from the calcium tablets.

Explanation / Answer

1)Each tablet contains 1.5 g CaC03

molar mass of CaC03 = 100 g

so

moles of CaC03 = mass / molar mass

moles of CaC03 = 1.5/100 = 15 x 10-3

CaC03 ---> Ca+2 + C03-

so

moles of Ca+2 = moles of CaC03 = 15 x 10-3

mass of Ca+2 = moles x molar mass

mass of Ca+2 = 15 x 10-3 x 40

mass of Ca+2 = 0.6 g

so

mass of Ca+2 in one calcium tablet is 0.6 g

2) from question 1

1.483 grams milk powder contains 0.0196 g

let y grams of milk powder contains 0.6 g

y = 0.6 x 1.483 / 0.0196

y = 45.40

so 45.40 g of milk powder is needed

no of cups = mass of milk powder / 22.7

no of cups = 45.40 / 22.7

no of cups = 2

3)
     no of tablets required to get 1 g Ca+2 = 1 / 0.6

no of tablets required = 1.66 = 2

cost of one tablet = 3.98 / 60 = 0.06633

cost of two tablets = 0.06633 x2 = $ 0.1326

so for 1 g of Ca+2 form calcium tablets = $ 0.1326

for 1 g of Ca+2 from milk powder = $ 1.75

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