Question

High concentrations of ammonia (NH3), nitrite ion, and nitrate ion in water can kill fish. Lethal concentrations of these species for rainbow trout are approximately 1.002 mg/L, 0.412 mg/L, and 1352.2 mg/L, respectively. Express these concentrations in molality units, assuming a solution density of 1.00 g/mL.a. ______m ammoniab. ______m nitrite ironc. ______m nitrate ion

Answer 1

Explanation:

It is known that molality is the number of moles present in kg of solution.

Mathematically,  Molality = $\frac{\text{no. of moles of solute}}{\text{mass of solvent in Kg}}$

The given data is as follows.

Molar mass of ammonia = 17 g/mol

Concentration = 1.002 mg/L = $\frac{0.001002 g/L}{17 g/mol}$

                        = $5.89 \times 10^{-4}$ mol/L

Also,    density = $\frac{1 g}{mL}$ = 1 kg/L

Therefore, molality will be calculated as follows.

        Molality = $\frac{5.89 \times 10^{-4} mol/L}{1 kg/L}$

                      = $5.89 \times 10^{-4} mol/kg$

And,

Molar mass of nitrite = 46 g/mol

Concentration = 0.387 mg/L = $\frac{0.000412 g/L}{46 g/mol}$

                        = $8.956 \times 10^{-6}$ mol/L

And, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality = $\frac{8.956 \times 10^{-6} mol/L}{1 kg/L}$

                          = $8.956 \times 10^{-6}$ mol/kg  

Now, Molar mass of nitarte = 62 g/mol

      Concentration = 1352.2 mg/L

                              = $\frac{1.3522 g/L}{62 g/mol}$

                              = 0.02181 mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality will be calculated as follows.

         Molality = $\frac{0.02181 mol/L}{1 kg/L}$

                       = 0.02181 mol/kg

Therefore, molality of given species is $5.89 \times 10^{-4} mol/kg$  for ammonia, $8.956 \times 10^{-6}$ mol/kg  for nitrite, and 0.02181 mol/kg for nitrate ion.