Question

How many grams of CaCO3 must be dissolved in water so that, after dilution to 500mL, the resulting solution contains 500ppm Ca? And what volume of the above stock should be diluted to 50mL so that the resulting solution contains 40ppm Ca?

Answer 1

we know that

ppm = mass of Ca in mg / volume of soluton (L)

so

500 = mass of Ca in mg / 0.5

mass of Ca in mg = 250

so

mass of Ca in g = 0.25

so

mass of Ca = 0.25 g

now

consider CaC03

we can see that

100 grams of CaC03 ---> 40 g of Ca

let y grams of CaC03 ---> 0.25 g of Ca

so

y = 0.25 x 100 / 40

y = 0.625

so

0.625 grams of CaC03 is required

2)

now

ppm = mass of Ca in mg / volume (L)

so

40 = mass of Ca in mg / 50 x 10-3

mass of Ca in mg = 2

now

this should come from the initial stock solution

so

ppm = mass of Ca in mg / volume (L)

500 = 2 / volume (L)

volume (L) = 4 x 10-3

volume (ml) = 4

so

to get 50 ml of 40 ppm Ca

4 ml of stock solution should be taken

and

water is added to make the solution 50 ml