Let’s tackle your questions one at a time, starting with the first one about osmotic pressure. To find the osmotic pressure of the resulting mixture when two solutions are combined, we can use the formula for osmotic pressure, which is given by the equation:
Understanding Osmotic Pressure
The osmotic pressure (π) of a solution can be calculated using the formula:
π = iCRT
Where:
- i = van 't Hoff factor (number of particles the solute dissociates into)
- C = molarity of the solution
- R = ideal gas constant (0.0821 L·atm/(K·mol))
- T = temperature in Kelvin
In your case, we have two solutions:
- Solution 1: 100 cc of 1.5% urea solution with an osmotic pressure of 6 atm
- Solution 2: 100 cc of 3.42% cane sugar solution with an osmotic pressure of 2.4 atm
Calculating the Total Osmotic Pressure
When mixing two solutions, the total osmotic pressure can be found by simply adding the individual osmotic pressures of the two solutions, assuming they behave ideally and do not interact significantly. Thus:
π_total = π_urea + π_sugar
Substituting the values:
π_total = 6 atm + 2.4 atm = 8.4 atm
So, the answer to your first question is a) 8.4 atm.
Determining Molecular Weight of a Solute
Now, let’s move on to your second question regarding the molecular weight of a non-volatile solute. We can use Raoult's Law, which states that the vapor pressure of a solvent is directly proportional to the mole fraction of the solvent in the solution.
The formula for the change in vapor pressure (ΔP) is:
ΔP = P° - P
Where:
- P° = vapor pressure of pure solvent
- P = vapor pressure of the solution
Given:
- ΔP = 2.25 x 10-1 mm
- P° = 17.5 mm
First, we can find the vapor pressure of the solution:
P = P° - ΔP = 17.5 mm - 0.225 mm = 17.275 mm
Calculating Mole Fraction and Molecular Weight
Next, we can find the mole fraction of the solute (X_solute) using the formula:
ΔP/P° = X_solute
Substituting the values:
2.25 x 10-1 mm / 17.5 mm = X_solute
X_solute = 0.012857
Now, we know that:
X_solute = n_solute / (n_solute + n_solvent)
Where:
- n_solute = moles of solute
- n_solvent = moles of solvent (water)
For 100 g of water, the number of moles of water (molar mass = 18 g/mol) is:
n_solvent = 100 g / 18 g/mol = 5.56 mol
Let’s denote the molar mass of the solute as M. The number of moles of solute can be expressed as:
n_solute = mass of solute / M = 25 g / M
Substituting into the mole fraction equation:
0.012857 = (25/M) / (25/M + 5.56)
Cross-multiplying and simplifying gives:
0.012857(25 + 5.56M) = 25
0.321425 + 0.071428M = 25
0.071428M = 24.678575
M = 24.678575 / 0.071428 ≈ 346.5 g/mol
After rounding, the closest answer from your options is c) 350.
In summary, the answers to your questions are:
- First question: a) 8.4 atm
- Second question: c) 350
