What is the osmotic pressure of a solution prepared by dissolving 5.80 g of CaCl2 in enough water to make 450. mL of solution at 24.7°C?
See the post by Lily just before your post. Post your work if you get stuck.
thanks
To find the osmotic pressure of a solution, you can use the formula:
π = i * M * R * T,
where:
π is the osmotic pressure (in atm),
i is the van't Hoff factor (which represents the number of particles produced per molecule or formula unit dissolved),
M is the molarity of the solution (in mol/L),
R is the ideal gas constant (0.0821 L*atm/(mol*K)),
T is the temperature (in Kelvin).
First, let's calculate the molarity (M) of the solution:
M = moles of solute / volume of solution in liters.
We have 5.80 g of CaCl2. To convert this to moles, we'll use the molar mass of CaCl2, which is 40.08 g/mol for Ca and 35.45 g/mol for Cl.
Molar mass of CaCl2 = (40.08 g/mol * 1 Ca) + (35.45 g/mol * 2 Cl) = 110.98 g/mol.
moles of CaCl2 = 5.80 g / 110.98 g/mol = 0.0522 mol.
Now, let's find the volume of the solution in liters:
450 mL = 450 mL * (1 L / 1000 mL) = 0.450 L.
Now, we can calculate the molarity (M):
M = 0.0522 mol / 0.450 L ≈ 0.116 M.
Next, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15.
T(K) = 24.7°C + 273.15 = 297.85 K.
The van't Hoff factor (i) for CaCl2 is 3 since it dissociates into 3 particles (Ca2+ and two Cl-) when dissolved.
Finally, we can calculate the osmotic pressure (π):
π = i * M * R * T.
π = 3 * 0.116 M * 0.0821 L*atm/(mol*K) * 297.85 K.
Calculating this will give you the osmotic pressure of the solution prepared by dissolving 5.80 g of CaCl2 in enough water to make 450. mL of solution at 24.7°C.