Question

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?

Answer 1

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

Solution : Given,

Moles of calcium =
`1.5* 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}* \text{Molar mass of calcium}=(1.5* 10^(-5)moles)* (40g/mole)=60* 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}* 1000}{\text{Volume of solution}}=(6000mg* 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

Answer 2

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