To calculate the osmotic pressure of a 0.85% aqueous solution of NaNO₃ that is 90% dissociated at 27°C, we can follow these steps:
### Step 1: Calculate the mass of NaNO₃ in the solution
Given that the solution is 0.85% NaNO₃, this means that there are 0.85 grams of NaNO₃ in 100 mL of solution.
### Step 2: Calculate the number of moles of NaNO₃
To find the number of moles, we use the formula:
\[
\text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
The molar mass of NaNO₃ (sodium nitrate) is approximately 85 g/mol. Therefore:
\[
\text{Number of moles of NaNO₃} = \frac{0.85 \, \text{g}}{85 \, \text{g/mol}} = 0.01 \, \text{mol}
\]
### Step 3: Adjust for dissociation
Since the solution is 90% dissociated, we must account for the dissociation of NaNO₃ into Na⁺ and NO₃⁻ ions. Each formula unit of NaNO₃ produces 2 ions upon dissociation. Therefore, the effective number of moles of particles (ions) is:
\[
\text{Effective moles} = 0.01 \, \text{mol} \times 2 \times 0.90 = 0.018 \, \text{mol}
\]
### Step 4: Convert the volume of the solution to liters
The volume of the solution is given as 100 mL, which is equivalent to:
\[
V = 100 \, \text{mL} = 0.1 \, \text{L}
\]
### Step 5: Convert temperature to Kelvin
The temperature in Celsius is given as 27°C. To convert this to Kelvin:
\[
T = 27 + 273 = 300 \, \text{K}
\]
### Step 6: Calculate the osmotic pressure using the formula
The formula for osmotic pressure (π) is given by:
\[
\pi = \frac{n}{V}RT
\]
Where:
- \( n \) = effective number of moles of solute (0.018 mol)
- \( V \) = volume of solution in liters (0.1 L)
- \( R \) = ideal gas constant (0.0821 atm K⁻¹ mol⁻¹)
- \( T \) = temperature in Kelvin (300 K)
Substituting the values:
\[
\pi = \frac{0.018 \, \text{mol}}{0.1 \, \text{L}} \times 0.0821 \, \text{atm K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}
\]
\[
\pi = 0.18 \times 0.0821 \times 300
\]
\[
\pi = 4.674 \, \text{atm}
\]
### Final Answer
The osmotic pressure of the solution is approximately **4.674 atm**.
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