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Suppose the mole number of Ca2+ ions in a 50 mL water sample is quantified as 1.5 × 10−5 mol. What is the concentration of Ca2+ ions in the water sample in ppm CaCO3?

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\(\frac{1.5x 10^{-5} mol Ca^{2+}}{50ml} ( \frac{1 mol CaC O_{3} }{1 mol Ca^{2+} } )( \frac{100,000 ppm}{1 molCaC O_{3}/L } ) \\ =0.0300234\)

Answer : The concentration of \(Ca^{2+}\) ions in the water sample in ppm is, 12000 ppm

Solution : Given,

Moles of calcium = \(1.5\times 10^{-5}moles\)

Molar mass of calcium = 40 g/mole

Volume of solution = 50 ml

First we have to calculate the mass of calcium.

\(\text{Mass of calcium}=\text{Moles of calcium}\times \text{Molar mass of calcium}=(1.5\times 10^{-5}moles)\times (40g/mole)=60\times 10^{-5}g=6000mg\)

(1 g = 1000 mg)

Now we have to calculate the concentration of calcium.

\(\text{Concentration of calcium}=\frac{\text{Mass of calcium}\times 1000}{\text{Volume of solution}}=\frac{6000mg\times 1000}{50ml}=12000mg/L=12000ppm\)

(1 ppm = 1 mg/L)

Therefore, the concentration of \(Ca^{2+}\) ions in the water sample in ppm is, 12000 ppm

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