Answer :
Answer :
The concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]
The concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]
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 = [tex]4\times 10^{-5}g/kg[/tex]
Conversion used : 1 g = 1000 mg
[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]
Thus, the concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]
Now we have to calculate concentration in molality of K.
[tex]\text{Concentration in molality}=\frac{\text{Concentration of K}}{\text{Molar mass of K}}[/tex]
Molar mass of K = 39 g/mol
Concentration of K = 1.30 mg/kg = [tex]1.3\times 10^{-3}g/kg[/tex]
Conversion used : 1 g = 1000 mg
[tex]\text{Concentration in molality}=\frac{1.3\times 10^{-3}g/kg}{39g/mol}[/tex]
[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]
Thus, the concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]