Osmotic pressure of `40%` (wt.`//`vol.) urea solution is `1.64 atm` and that of `3.42%` (wt.`//`vol.) cane sugar is `2.46 atm`. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is:

Options

A. `1.64 atm`
B. `2.46 atm`
C. `4.10 atm`
D. `2.05 atm`

Explanation

To find the osmotic pressure of the resulting solution when equal volumes of a 40% (wt/vol.) urea solution and a 3.42% (wt/vol.) cane sugar solution are mixed, we can follow these steps:

### Step 1: Determine the initial concentrations of both solutions.
1. **Urea Solution:**
- Given: 40% (wt/vol.) urea solution means 40 grams of urea in 100 mL of solution.
- Molar mass of urea (NH₂CONH₂) = 60 g/mol.
- Moles of urea = mass/molar mass = 40 g / 60 g/mol = 2/3 mol.
- Concentration (C1) = moles/volume (in L) = (2/3 mol) / (0.1 L) = 20/3 mol/L.

2. **Cane Sugar Solution:**
- Given: 3.42% (wt/vol.) cane sugar 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 = mass/molar mass = 3.42 g / 342 g/mol = 1/100 mol.
- Concentration (C2) = moles/volume (in L) = (1/100 mol) / (0.1 L) = 1/10 mol/L.

### Step 2: Calculate the concentrations after mixing.
When equal volumes of both solutions are mixed, the total volume doubles, so the concentrations will be halved:
- New concentration of urea (C1') = (20/3 mol/L) / 2 = 10/3 mol/L.
- New concentration of cane sugar (C2') = (1/10 mol/L) / 2 = 1/20 mol/L.

### Step 3: Calculate the osmotic pressures of the individual solutions.
Using the formula for osmotic pressure (π = iCRT):
- For urea (i = 1):
- π1 = i1 * C1' * RT = 1 * (10/3 mol/L) * RT = (10/3)RT.

- For cane sugar (i = 1):
- π2 = i2 * C2' * RT = 1 * (1/20 mol/L) * RT = (1/20)RT.

### Step 4: Find the total osmotic pressure of the resulting solution.
The total osmotic pressure (π_total) when both solutions are mixed:
- π_total = π1 + π2 = (10/3)RT + (1/20)RT = RT[(10/3) + (1/20)].

### Step 5: Find a common denominator and simplify.
- Common denominator for 3 and 20 is 60:
- (10/3) = (200/60) and (1/20) = (3/60).
- π_total = RT[(200/60) + (3/60)] = RT(203/60).

### Step 6: Determine RT using the given osmotic pressures.
From the original solutions:
1. For urea: π1 = 1.64 atm = (20/3)RT → RT = (1.64 * 3) / 20 = 0.246 atm·L/mol.
2. For cane sugar: π2 = 2.46 atm = (1/10)RT → RT = (2.46 * 10) = 24.6 atm·L/mol.

### Step 7: Calculate the osmotic pressure of the resulting solution.
Using the average RT value:
- π_total = (203/60) * RT = (203/60) * 0.246 atm·L/mol.
- π_total = 2.05 atm.

### Final Answer:
The osmotic pressure of the resulting solution is approximately **2.05 atm**.