Question

a) A water sample (density=1.00g/mL, S=0.28g/kg) contains Ca(2+) at a concentration of 42 mg/kg. Calculate the molarity of the ion.

Answer 1

Answer : The molarity of the ion is
`1.05* 10^(-3)M`

Explanation : Given,

Density of sample = 1.00 g/mL

Concentration of
`Ca^(2+)` = 42 mg/kg

First we have to calculate the volume of sample.

Let us assume that the mass of sample be 1 kg or 1000 g.

`Density=(Mass)/(Volume)`

`1.00g/mL=(1000g)/(Volume)`

`Volume=(1000g)/(1.00g/mL)`

`Volume=1000mL=1L` (1 L = 1000 mL)

Now we have to calculate the moles of
`Ca^(2+)`

As we are given that,

Mass of
`Ca^(2+)` in 1 kg of sample = 42 mg = 0.042 g

Molar mass of Ca = 40 g/mole

`\text{Moles of }Ca^(2+)=\frac{\text{Mass of }Ca^(2+)}{\text{Molar mass of }Ca^(2+)}=(0.042g)/(40g/mol)=1.05* 10^(-3)mol`

Now we have to calculate the molarity of the ion.

`\text{Molarity}=\frac{\text{Moles of }Ca^(2+)}{\text{Volume of sample}}`

`\text{Molarity}=(1.05* 10^(-3)mol)/(1L)=1.05* 10^(-3)mol/L=1.05* 10^(-3)M`

Therefore, the molarity of the ion is
`1.05* 10^(-3)M`