To find the osmotic pressure of the solution containing urea and cane sugar, we can follow these steps:
### Step 1: Calculate the number of moles of urea
- The concentration of urea is given as 3.4% (w/v), which means there are 3.4 grams of urea in 100 ml of solution.
- To find the number of moles of urea, we use the formula:
\[
\text{Moles of urea} = \frac{\text{mass of urea (g)}}{\text{molar mass of urea (g/mol)}}
\]
Given that the molar mass of urea is 60 g/mol:
\[
\text{Moles of urea} = \frac{3.4 \, \text{g}}{60 \, \text{g/mol}} = 0.0567 \, \text{moles}
\]
### Step 2: Calculate the number of moles of cane sugar
- The concentration of cane sugar is given as 1.6% (w/v), which means there are 1.6 grams of cane sugar in 100 ml of solution.
- Since we are using 50 ml of this solution, the mass of cane sugar in 50 ml is:
\[
\text{Mass of cane sugar} = \frac{1.6 \, \text{g}}{100 \, \text{ml}} \times 50 \, \text{ml} = 0.8 \, \text{g}
\]
- To find the number of moles of cane sugar, we use the formula:
\[
\text{Moles of cane sugar} = \frac{\text{mass of cane sugar (g)}}{\text{molar mass of cane sugar (g/mol)}}
\]
Given that the molar mass of cane sugar is 342 g/mol:
\[
\text{Moles of cane sugar} = \frac{0.8 \, \text{g}}{342 \, \text{g/mol}} = 0.00234 \, \text{moles}
\]
### Step 3: Calculate the total number of moles
- Total moles in the solution:
\[
\text{Total moles} = \text{Moles of urea} + \text{Moles of cane sugar} = 0.0567 + 0.00234 = 0.05904 \, \text{moles}
\]
### Step 4: Calculate the total volume of the solution in liters
- The total volume of the solution is:
\[
\text{Total volume} = 100 \, \text{ml} + 50 \, \text{ml} = 150 \, \text{ml} = 0.150 \, \text{liters}
\]
### Step 5: Calculate the molarity of the solution
- Molarity (C) is given by:
\[
C = \frac{\text{Total moles}}{\text{Total volume (L)}} = \frac{0.05904 \, \text{moles}}{0.150 \, \text{L}} = 0.3936 \, \text{M}
\]
### Step 6: Calculate the osmotic pressure using the formula
- The osmotic pressure (π) can be calculated using the formula:
\[
\pi = CRT
\]
where:
- \( C = 0.3936 \, \text{M} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 27^\circ C = 300 \, \text{K} \)
Substituting the values:
\[
\pi = 0.3936 \times 0.0821 \times 300
\]
Calculating:
\[
\pi = 0.3936 \times 24.63 \approx 9.69 \, \text{atm}
\]
### Final Answer
The osmotic pressure of the solution is approximately **9.7 atm**.
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