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 $1.05\times 10^{-3}M$

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.

$Density=\frac{Mass}{Volume}$

$1.00g/mL=\frac{1000g}{Volume}$

$Volume=\frac{1000g}{1.00g/mL}$

$Volume=1000mL=1L$        (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

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

Now we have to calculate the molarity of the ion.

$\text{Molarity}=\frac{\text{Moles of }Ca^{2+}}{\text{Volume of sample}}$

$\text{Molarity}=\frac{1.05\times 10^{-3}mol}{1L}=1.05\times 10^{-3}mol/L=1.05\times 10^{-3}M$

Therefore, the molarity of the ion is $1.05\times 10^{-3}M$