.361 g of a non-volatile solute is dissolved in 160.0 g of water. The solute does not react with water nor dissociate in solution. Assume that the resulting solution displays ideal Raoult's law behaviour. At 90 oC the vapour pressure of the solution is 5.1701×102 torr. (The vapour pressure of pure water at 90 oC is 5.2580×102 torr.) Calculate the molar mass of the solute (in g).
I keep getting 541.1g/mol. what am I doing wrong? Help me please
Now suppose, instead, that the solute in problem 6 is volatile and that the pure solute displays, at 90 oC, a vapour pressure of 52.580 torr. Calculate the molar mass (in g) for this case.
This is the second part to this question
Please do not switch names -- especially on the same post!!!!!
To determine the molar mass of the solute, we can apply Raoult's law, which states that the vapor pressure above a solution is directly proportional to the mole fraction of the solvent present in the solution.
First, let's calculate the mole fraction of the solvent (water) in the solution. The mole fraction (X) of water can be calculated using the equation:
X_water = moles of water / total moles of solute and solvent
To find the moles of water, we divide the mass of water by its molar mass (18.015 g/mol):
moles of water = mass of water / molar mass of water
= 160.0 g / 18.015 g/mol
≈ 8.8802 mol
Next, we need to calculate the mole fraction of solute (non-volatile solute). Since the solute is non-volatile and does not dissociate in the solution, the mole fraction of the solute is equal to 1 - mole fraction of water:
X_solute = 1 - X_water
Now, we can calculate the mole fraction of the solute. Given that the mole fraction of the solute is directly proportional to the vapor pressure of the solution (according to Raoult's law), we can write it as:
X_solute = P_solute / P_solution
Given that the vapor pressure of the water is 5.2580×102 torr and the vapor pressure of the solution is 5.1701×102 torr, we can substitute these values into the equation to solve for X_solute:
X_solute = (5.1701×102 torr) / (5.2580×102 torr)
Calculate this division to find X_solute:
X_solute ≈ 0.9838
Now, we can substitute the value of X_solute into the equation for mole fraction of the solute:
0.9838 = 1 - X_water
Solving for X_water:
X_water ≈ 1 - 0.9838
X_water ≈ 0.0162
Now, let's find the moles of water by multiplying the mole fraction by the total moles of the solution:
moles of water = X_water * (total moles of solute and solvent)
= 0.0162 * 8.8802 mol
≈ 0.1439 mol
Since we know the mass of the solute (0.361 g) and the moles of water (0.1439 mol), we can calculate the moles of solute by subtracting the moles of water from the total moles of the solution:
moles of solute = total moles of solute and solvent - moles of water
= 8.8802 mol - 0.1439 mol
= 8.7363 mol
Finally, we can calculate the molar mass of the solute by dividing the mass of the solute by the moles of the solute:
molar mass of solute = mass of solute / moles of solute
= 0.361 g / 8.7363 mol
≈ 0.041 g/mol
The molar mass of the solute is approximately 0.041 g/mol, not 541.1 g/mol. Double-check your calculations to identify any errors.