100 cc of 1.5% solution of urea is found to have an osmotic pressure of 6.0 atm and 100 cc of 3.42% solution of cane sugar is found to have an osmotic pressure of 2.4 atm. If two solutions are mixed, the osmotic pressure of the resulting solution will be

Options

A. 8.4 atm
B. 4.2 atm
C. 16.8 atm
D. 2.1 atm

Explanation

To find the osmotic pressure of the resulting solution when mixing two solutions, we can use the formula for osmotic pressure, which is given by:

\[ \pi = iCRT \]

Where:
- \(\pi\) = osmotic pressure
- \(i\) = van 't Hoff factor (which is 1 for urea and cane sugar since they do not dissociate in solution)
- \(C\) = molarity of the solution
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin (assumed constant for both solutions)

### Step 1: Calculate the molarity of each solution

1. **Urea Solution:**
- Given: 1.5% w/v solution of urea.
- This means 1.5 grams of urea in 100 mL of solution.
- Molar mass of urea (NH₂CONH₂) = 60 g/mol.
- Moles of urea = \(\frac{1.5 \text{ g}}{60 \text{ g/mol}} = 0.025 \text{ mol}\).
- Molarity (C) = \(\frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.025 \text{ mol}}{0.1 \text{ L}} = 0.25 \text{ M}\).

2. **Cane Sugar Solution:**
- Given: 3.42% w/v solution of cane sugar.
- This means 3.42 grams of cane sugar in 100 mL of solution.
- Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 342 g/mol.
- Moles of cane sugar = \(\frac{3.42 \text{ g}}{342 \text{ g/mol}} = 0.01 \text{ mol}\).
- Molarity (C) = \(\frac{0.01 \text{ mol}}{0.1 \text{ L}} = 0.1 \text{ M}\).

### Step 2: Calculate the osmotic pressure for each solution

1. **Osmotic Pressure of Urea Solution:**
- Using the formula \(\pi = iCRT\):
- \(\pi_{\text{urea}} = (1)(0.25 \text{ M})(0.0821 \text{ L·atm/(K·mol)})(T)\).
- Since we are not given temperature, we will assume it is constant and can be canceled out later.

2. **Osmotic Pressure of Cane Sugar Solution:**
- \(\pi_{\text{sugar}} = (1)(0.1 \text{ M})(0.0821 \text{ L·atm/(K·mol)})(T)\).

### Step 3: Calculate the total osmotic pressure when mixed

- When the two solutions are mixed, the total volume becomes 200 mL (0.2 L).
- The total moles of solute remain the same, so we can calculate the new concentrations:

1. **Total moles of solute:**
- Moles from urea = 0.025 mol
- Moles from cane sugar = 0.01 mol
- Total moles = \(0.025 + 0.01 = 0.035 \text{ mol}\).

2. **New concentration after mixing:**
- Total volume = 0.2 L.
- New concentration \(C_{\text{total}} = \frac{0.035 \text{ mol}}{0.2 \text{ L}} = 0.175 \text{ M}\).

3. **Calculate the total osmotic pressure:**
- \(\pi_{\text{total}} = (1)(0.175 \text{ M})(0.0821 \text{ L·atm/(K·mol)})(T)\).

### Final Result

Since we are interested in the osmotic pressure, we can summarize:
- The resulting osmotic pressure will be proportional to the new concentration, which is \(0.175 \text{ M}\).