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If 0.4575 g of CaCO3 were dissolved and diluted in a 300.00 mL volumetric flask (assume...

If 0.4575 g of CaCO3 were dissolved and diluted in a 300.00 mL volumetric flask (assume the density of the solution is 1.00 g/mL), what would the calcium (Ca2+) concentration be in a) molarity and b)ppm?

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Answer #1

Mass = volume * density

= 300 *1

= 300 grams

Ppb = (mass of solute / mass of Solution ) *10^6

= (0.4575 / 300 ) *10^6

= 1525

CaCO3 = Ca2+(aq) + CO3 2- (aq)

Moles of CaCO3 = moles of Ca2+

= Mass / molar mass of CaCO3

= 0.4575 / 100

= 4.575*10^-3

Molarity = moles / volume (in litre)

= (4.575*10^-3 / 0.300)

= 0.01525

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