Question

A solution of lead (ii) ions in water has a concentration of 10^4ppm. What is the osmotic pressure of the solution at 300k? assume density of 1.00g/ml

Answer 1

Let us take 100 mL solution

mass of solution = (volume of solution) * (density of solution)

mass of solution = (100 mL) * (1.00 g/mL)

mass of solution = 100 g

mass of lead = (concentration of lead) * (mass of solution)

mass of lead = (10⁴ ppm) * (100 g)

mass of lead = (10⁴ / 10⁶) * (100 g)

mass of lead = 1 g

moles of lead = (mass of lead) / (molar mass of lead)

moles of lead = (1 g) / (207.2 g/mol)

moles of lead = 4.83 x 10^-3 mol

Concentration of lead = (moles of lead) / (volume of solution is Liters)

Concentration of lead = (4.83 x 10^-3 mol) / (0.100 L)

Concentration of lead = 0.0483 M

osmotic pressure of solution = (Concentration of lead) * (R) * (Temperature)

where R = constant = 0.0821 L-atm/mol-K

osmotic pressure of solution = (0.0483 M) * (0.0821 L-atm/mol-K) * (300 K)

osmotic pressure of solution = 1.188 atm