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

Step-by-step 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 =
`(0.001002 g/L)/(17 g/mol)`

=
`5.89 * 10^(-4)` mol/L

Also, density =
`(1 g)/(mL)` = 1 kg/L

Therefore, molality will be calculated as follows.

Molality =
`(5.89 * 10^(-4) mol/L)/(1 kg/L)`

=
`5.89 * 10^(-4) mol/kg`

And,

Molar mass of nitrite = 46 g/mol

Concentration = 0.387 mg/L =
`(0.000412 g/L)/(46 g/mol)`

=
`8.956 * 10^(-6)` mol/L

And, density =
`(1 g)/(mL)` = 1 kg/L

Hence, molality =
`(8.956 * 10^(-6) mol/L)/(1 kg/L)`

=
`8.956 * 10^(-6)` mol/kg

Now, Molar mass of nitarte = 62 g/mol

Concentration = 1352.2 mg/L

=
`(1.3522 g/L)/(62 g/mol)`

= 0.02181 mol/L

Also, density =
`(1 g)/(mL)` = 1 kg/L

Hence, molality will be calculated as follows.

Molality =
`(0.02181 mol/L)/(1 kg/L)`

= 0.02181 mol/kg

Therefore, molality of given species is
`5.89 * 10^(-4) mol/kg` for ammonia,
`8.956 * 10^(-6)` mol/kg for nitrite, and 0.02181 mol/kg for nitrate ion.