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, 0.436 mg/L, and 1362.4 mg/L, respectively. Express these concentrations in molality units, assuming a solution density of 1.00 g/mL.

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, 0.436 mg/L, and 1362.4 mg/L, respectively. Express these concentrations in molality units, assuming a solution density of 1.00 g/mL.

Answer 1

All of these are done the same way. I'll do NH3.

1.002 mg/L and you want to convert to mols/kg solvent.
You have 1000 mL. The density is 1.00 g/mL so 1,000 mL has a mass of 1,000 grams. That is mass H2O + mass NH3 = 1,000 grams. The mass of NH3 is 1.002 mg or 0.001002 g. So mass H2O = 1,000 - 0.001002 = ? which is essentially 1,000 grams or 1 kg solvent. That will be true for nitrite ion too but probably will not be true for nitrate.
Then mols NH3 = grams/molar mass = 0.001002/17 = ?
Then molality = mols/kg solvent = ?
Post your work if you get stuck.

Answer 2

To convert the concentrations from mg/L to molality units, we need to consider the molar mass of each compound.

The molar mass of ammonia (NH3) is 17.03 g/mol, nitrite ion (NO2-) is 46.01 g/mol, and nitrate ion (NO3-) is 62.00 g/mol.

First, let's convert the concentrations of ammonia, nitrite ion, and nitrate ion from mg/L to g/L:

For ammonia (NH3):
1.002 mg/L * (1 g / 1000 mg) = 0.001002 g/L

For nitrite ion (NO2-):
0.436 mg/L * (1 g / 1000 mg) = 0.000436 g/L

For nitrate ion (NO3-):
1362.4 mg/L * (1 g / 1000 mg) = 1.3624 g/L

Next, let's convert the concentrations from g/L to molality using the density of the solution.

Since the density of the solution is 1.00 g/mL, we can convert the concentrations directly to molality by dividing by the molar masses:

For ammonia (NH3):
0.001002 g/L / 17.03 g/mol = 0.0000589 mol/L

For nitrite ion (NO2-):
0.000436 g/L / 46.01 g/mol = 0.00000948 mol/L

For nitrate ion (NO3-):
1.3624 g/L / 62.00 g/mol = 0.02199 mol/L

Finally, convert the mol/L to molality units by dividing by the solution density:

For ammonia (NH3):
0.0000589 mol/L / 1.00 g/mL = 0.0000589 mol/g

For nitrite ion (NO2-):
0.00000948 mol/L / 1.00 g/mL = 0.00000948 mol/g

For nitrate ion (NO3-):
0.02199 mol/L / 1.00 g/mL = 0.02199 mol/g

Therefore, the concentrations in molality units for ammonia, nitrite ion, and nitrate ion in water are approximately:
- Ammonia (NH3): 0.0000589 mol/g
- Nitrite ion (NO2-): 0.00000948 mol/g
- Nitrate ion (NO3-): 0.02199 mol/g

Answer 3

To express the concentrations of ammonia, nitrite ion, and nitrate ion in molality units, we need to convert the given concentrations from mass per volume (mg/L) to moles per kilogram (mol/kg).

First, let's convert the concentration of ammonia (NH3) from mg/L to g/L:

1.002 mg/L = 1.002 g/L

Given that the density of the solution is 1.00 g/mL, we can assume that 1.00 g of the solution has a volume of 1 mL, which is equal to 1 L. Therefore, 1.002 g/L can be considered as 1.002 g/kg.

Now, let's calculate the molar mass of ammonia:

NH3 = 1(N) + 3(H) = 14.0067 g/mol + 3(1.00784 g/mol) ≈ 17.03052 g/mol

Next, we can calculate the molality of ammonia:

Molality (m) = (mass of solute in moles) / (mass of solvent in kg)

Mass of solute (ammonia) = 1.002 g
Mass of solvent (water) = 1000 g (because 1 L of water = 1000 g)

Mass of solute in moles = 1.002 g / 17.03052 g/mol ≈ 0.05888 mol

Molality (ammonia) = 0.05888 mol / 1.000 kg ≈ 0.05888 mol/kg

Now let's repeat this process for the nitrite ion (NO2-) and nitrate ion (NO3-).

For the nitrite ion:

0.436 mg/L = 0.436 g/L = 0.436 g/kg

The molar mass of nitrite ion (NO2-) = 14.0067 g/mol + 2(16.00 g/mol) ≈ 46.0055 g/mol

Mass of solute in moles = 0.436 g / 46.0055 g/mol ≈ 0.0094927 mol

Molality (nitrite ion) = 0.0094927 mol / 1.000 kg ≈ 0.0094927 mol/kg

For the nitrate ion:

1362.4 mg/L = 1362.4 g/L = 1362.4 g/kg

The molar mass of nitrate ion (NO3-) = 14.0067 g/mol + 3(16.00 g/mol) ≈ 62.0049 g/mol

Mass of solute in moles = 1362.4 g / 62.0049 g/mol ≈ 21.9995 mol

Molality (nitrate ion) = 21.9995 mol / 1.000 kg ≈ 21.9995 mol/kg

So, we have:

Molality (ammonia) ≈ 0.05888 mol/kg
Molality (nitrite ion) ≈ 0.0094927 mol/kg
Molality (nitrate ion) ≈ 21.9995 mol/kg

Therefore, the concentrations of ammonia, nitrite ion, and nitrate ion in molality units are approximately 0.05888 mol/kg, 0.0094927 mol/kg, and 21.9995 mol/kg, respectively.