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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

User Inarilo
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1 Answer

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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.

User Qi Fan
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