The osmotic pressure at 17°C of an aqueous solution containing 1.75 g…

The correct answer is option 3. 0.81.

To find the osmotic pressure (\(\pi\)) of the solution, we can use the formula:

\(\pi = i \cdot M \cdot R \cdot T \)

Where:

\( i \) is the van 't Hoff factor (which is 1 for sucrose),

\( M \) is the molarity of the solution,

\( R \) is the ideal gas constant (0.08206 L atm K\(^{-1}\) mol\(^{-1}\)),

\( T \) is the temperature in Kelvin.

First, let's find the molarity (\( M \)) of the solution using the given data:

\(M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

Given:

Mass of sucrose (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)) = 1.75 g,

Volume of solution = 150 mL,

Molar mass of sucrose = 342 g/mol.

\(\text{moles of sucrose} = \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} \)

\(\text{moles of sucrose} = \frac{1.75 \, \text{g}}{342 \, \text{g/mol}} = 0.005116 \, \text{mol} \)

\(\text{Volume of solution} = 150 \, \text{mL} = 0.150 \, \text{L} \)

\(M = \frac{0.005116 \, \text{mol}}{0.150 \, \text{L}} = 0.034107 \, \text{M} \)

Now, we have the molarity of the solution. We'll use this value along with the given temperature to calculate the osmotic pressure.

Given temperature (\( T \)) = 17°C = 17 + 273.15 = 290.15 K.

Now, we can plug in the values into the formula:

\(\pi = (1) \times (0.034107 \, \text{M}) \times (0.08206 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (290.15 \, \text{K}) \)
\(\pi \approx 0.83 \, \text{atm}\)

Therefore, the osmotic pressure of the solution is approximately 0.83 atm.

None of the given options exactly match the calculated value, but the closest option is 0.81 atm.