When 3.89 g of a nonelectrolyte solute is dissolved in water to make 835 mL of solution at 26 °C, the solution exerts an osmotic pressure of 825 torr.
What is the molar concentration of the solution?
How many moles of solute are in the solution?
What is the molar mass of the solute?
pi = nRT
Substitute and solve for n = mols
Then M = mols/L solution
# mols = M x L = ?
molar mass = grams/mols.
To find the molar concentration of the solution, we can use the formula:
Molar concentration (C) = moles of solute (n) / volume of solution (V)
To find the moles of solute, we can use the formula:
n = (osmotic pressure (Pi) * V) / (RT)
Where:
- osmotic pressure (Pi) is given as 825 torr
- volume of solution (V) is given as 835 mL, but we need to convert it to liters (1 L = 1000 mL)
- R is the ideal gas constant (0.0821 L*atm/(mol*K))
- T is the temperature in Kelvin (26 °C = 26 + 273 = 299 K)
First, let's convert the volume of the solution to liters:
V = 835 mL / 1000 mL/L = 0.835 L
Now we can substitute the given values into the equation for moles of solute:
n = (825 torr * 0.835 L) / (0.0821 L*atm/(mol*K) * 299 K)
n ≈ 26.6 mol
So, there are approximately 26.6 moles of solute in the solution.
Next, to find the molar concentration of the solution, we will substitute the values of moles of solute and volume of solution into the equation for molar concentration:
C = 26.6 mol / 0.835 L
C ≈ 31.80 M
Therefore, the molar concentration of the solution is approximately 31.80 M.
To find the molar mass of the solute, we can use the formula:
Molar mass (M) = mass of solute (m) / moles of solute (n)
We are given the mass of the solute as 3.85 g and we calculated the moles of solute as approximately 26.6 mol. Substituting these values into the formula, we get:
M = 3.85 g / 26.6 mol
M ≈ 0.15 g/mol
Therefore, the molar mass of the solute is approximately 0.15 g/mol.