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, 0.387 mg/L, and 1352.2 mg/L, respectively. Express these concentrations in molality units, assuming a solution density of 1.00 g/mL.

______m ammonia

______m nitrite iron

______m nitrate ion

Answer 1

Molality = No. of moles of solute / mass of Solvent in Kg

Given

Ammonia:

Molar mass = 17 g/mol

Concentration = 1.002 mg/L = 0.001002 g/L / 17g/mol = 5.89 * 10^-4 mol/L

Given density = 1g/mL = 1kg/L

so Molality = 5.89 * 10^-4 mol/L /1kg/L = 5.89 * 10^-4 mol/kg Answer

Nitrite:

Molar mass = 46 g/mol

Concentration = 0.387 mg/L = 0.000387 g/L / 46 g/mol = 8.413 * 10^-6 mol/L

Given density = 1g/mL = 1kg/L

so Molality = 8.413 * 10^-6 mol/L /1kg/L = 8.413 * 10^-6 mol/kg Answer

Nitrate:

Molar mass = 62 g/mol

Concentration = 1352.2 mg/L = 1.3522 g/L / 62 g/mol = 0.02181 mol/L

Given density = 1g/mL = 1kg/L

so Molality = 0.02181 mol/L /1kg/L = 0.02181 mol/kg Answer