Question

The concentrations of Fe and K in a sample of riverwater are 0.0400 mg/kg and 1.30 mg/kg, respectively. Express the concentration in molality.

Answer 1

Answer :

The concentration in molality of Fe is, $7.1\times 10^{-7}mol/kg$

The concentration in molality of K is, $3.3\times 10^{-5}mol/kg$

Explanation:

First we have to calculate concentration in molality of Fe.

Molar mass of Fe = 56 g/mol

Concentration of Fe = 0.0400 mg/kg = $4\times 10^{-5}g/kg$

Conversion used : 1 g = 1000 mg

$\text{Concentration in molality}=7.1\times 10^{-7}mol/kg$

Thus, the concentration in molality of Fe is, $7.1\times 10^{-7}mol/kg$

Now we have to calculate concentration in molality of K.

$\text{Concentration in molality}=\frac{\text{Concentration of K}}{\text{Molar mass of K}}$

Molar mass of K = 39 g/mol

Concentration of K = 1.30 mg/kg = $1.3\times 10^{-3}g/kg$

Conversion used : 1 g = 1000 mg

$\text{Concentration in molality}=\frac{1.3\times 10^{-3}g/kg}{39g/mol}$

$\text{Concentration in molality}=7.1\times 10^{-7}mol/kg$

Thus, the concentration in molality of K is, $3.3\times 10^{-5}mol/kg$