The amount of lead in drinking water is 0.025 mg/L. Express this concentration in molality. {molar mass of lead is 207.2 g/mol and the density of water is 1.00 g/cm.}
0.025 mg = 25E-3 g Pb.
mols Pb = grams Pb/atomic mass Pb.
1000 mL H2O x 1.00 g/mL = 1000 g = 1 kg.
molality = mols Pb/kg H2O
1.2E-4 mols/kg
To express the lead concentration in drinking water in molality, we need to convert the given concentration from mg/L to moles of lead per kilogram of water.
First, let's convert the concentration from mg/L to g/L:
0.025 mg/L * (1 g/1000 mg) = 0.025 g/L
Next, we need to convert the grams of lead to moles. To do this, we divide by the molar mass of lead:
0.025 g/L / 207.2 g/mol = 0.0001206 mol/L
To express the concentration in molality, we need to relate moles of solute (lead) to the mass of the solvent (water). The given density of water will be useful for this.
Density = mass/volume
Since the density of water is 1.00 g/cm³, which is equivalent to 1000 g/L, we can calculate the mass of water in kilograms.
Mass of water = Density * Volume
Mass of water = 1000 g/L * (1 L/1000 mL) * (1000 mL/1 cm³) * (1 cm³/1 mL) * (1 kg/1000 g)
Mass of water = 1 kg
Therefore, to convert the lead concentration from mol/L to mol/kg, we simply divide it by the mass of water:
0.0001206 mol/L / 1 kg = 0.0001206 mol/kg
So, the concentration of lead in drinking water expressed in molality is 0.0001206 mol/kg.