What is the osmotic pressure at 25°C of a solution containing 0.50 mol/L of Al(NO3)3?
pi in atm = i*MRT
i is 4 for Al(NO3)3; i.e., 1 for Al + 3 for NO3^-
M is given
R = 0.08206
T is 298 K
To calculate the osmotic pressure of a solution, you can use the formula:
π = nRT
Where:
π is the osmotic pressure (in Pascal or atmospheres)
n is the concentration of the solute in moles per liter (mol/L)
R is the ideal gas constant (8.314 J/(mol·K) or 0.08206 L·atm/(mol·K))
T is the temperature in Kelvin (K)
First, you have the concentration of the solution, which is 0.50 mol/L. Since Al(NO3)3 dissociates to cations and anions when dissolved in water, it will contribute a total of 4 particles (one aluminum ion and three nitrate ions) per formula unit.
Thus, the concentration of the solute in moles per liter (n) is given by:
n = (0.50 mol/L) × 4 = 2.00 mol/L
Next, convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Now, you can substitute the values into the formula:
π = (2.00 mol/L) × (0.08206 L·atm/(mol·K)) × (298.15 K)
Calculating the expression will give you the osmotic pressure of the solution.