69.7 grams of a solute with a molecular mass of 2790 grams are dissolved in enough water to make 1.00 dm3 of solution at 20 degrees C. What is the osmotic pressure of the soluton?
60.8
To calculate the osmotic pressure of a solution, you can use the formula:
π = (n/V)RT
Where:
π represents the osmotic pressure of the solution.
n is the number of moles of solute.
V is the volume of the solution.
R is the ideal gas constant (0.0821 L·atm/(mol·K)).
T is the temperature in Kelvin.
To find the number of moles of solute, you need to divide the mass of the solute by its molar mass:
n = (m/M)
Where:
n is the number of moles.
m is the mass of the solute.
M is the molar mass of the solute.
Now let's calculate step by step.
1. Calculate the number of moles of solute:
m = 69.7 grams
M = 2790 grams/mol
n = (69.7 g) / (2790 g/mol)
2. Convert the volume to liters:
1.00 dm^3 = 1.00 L
3. Convert the temperature to Kelvin:
20 degrees C = 20 + 273.15 K = 293.15 K
4. Substitute the values into the osmotic pressure formula:
π = (n/V)RT
= [(69.7 g) / (2790 g/mol)] / (1.00 L) * (0.0821 L·atm/(mol·K)) * (293.15 K)
5. Finally, calculate the osmotic pressure:
π = (69.7/2790) * (0.0821) * (293.15) atm
Now you can calculate π to find the osmotic pressure of the solution.