Answer :
1.84 atm is the osmotic pressure of an aqueous solution of 1.64 g Ca(NO₃)₂ in water at 25°C
To calculate the osmotic pressure of an aqueous solution, we need to follow these steps:
Determine the molar mass of Ca(NO₃)₂:
- Ca = 40.08 g/mol
- N = 14.01 g/mol (2 atoms)
- O = 16.00 g/mol (6 atoms)
- Molar mass = 40.08 + (2 * 14.01) + (6 * 16.00) = 164.1 g/mol
Determine the number of moles of Ca(NO₃)₂ in the solution:
- Given mass = 1.64 g
- Moles of Ca(NO₃)₂ = [tex]\frac{1.64 \text{ g}}{164.1 \text{ g/mol}} \approx 0.01 \text{ mol}[/tex]
Determine the number of moles of ions in the solution:
- Ca(NO₃)₂ dissociates into 1 Ca²⁺ ion and 2 NO₃⁻ ions.
- Total ions = 3 ions
Determine the molarity of the ions:
- Volume of solution = 400 mL = 0.400 L
- Molarity (M) = [tex]\frac{0.01 \text{ mol}}{0.400 \text{ L}} = 0.025 \text{ M}[/tex]
- Since each formula unit of Ca(NO₃)₂ produces 3 ions, the total molarity of ions is [tex]0.025 \text{ M} \times 3 = 0.075 \text{ M}[/tex]
Determine the osmotic pressure using the formula:
- Osmotic pressure (π) = MRT
- R (universal gas constant) = 0.0821 L·atm/(mol·K)
- T = 25°C = 298 K
- π = 0.075 M * 0.0821 L·atm/(mol·K) * 298 K
- π ≈ 1.84 atm
