The osmotic pressure at `17^(@)C` of aqueous solution containing `1.75g` of sucrose per `150ml`solution is

Options

A. 0.8 atm
B. 0.08 atm
C. 8.1 atm
D. 9.1 atm

Explanation

To find the osmotic pressure of the aqueous solution containing 1.75 g of sucrose per 150 mL, we will use the formula for osmotic pressure:

\[ \pi = \frac{nRT}{V} \]

Where:
- \(\pi\) = osmotic pressure
- \(n\) = number of moles of solute
- \(R\) = gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin
- \(V\) = volume of the solution in liters

### Step 1: Convert the mass of sucrose to moles
To find the number of moles (\(n\)), we use the formula:

\[ n = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of sucrose (C₁₂H₂₂O₁₁) is approximately 342 g/mol.

\[ n = \frac{1.75 \text{ g}}{342 \text{ g/mol}} \approx 0.00511 \text{ mol} \]

### Step 2: Convert the volume from mL to L
The volume of the solution is given as 150 mL. We need to convert this to liters:

\[ V = \frac{150 \text{ mL}}{1000} = 0.150 \text{ L} \]

### Step 3: Convert the temperature from Celsius to Kelvin
The temperature is given as 17°C. To convert this to Kelvin, we add 273.15:

\[ T = 17 + 273.15 = 290.15 \text{ K} \]

### Step 4: Substitute the values into the osmotic pressure formula
Now we can substitute the values into the osmotic pressure formula:

\[ \pi = \frac{nRT}{V} \]

Substituting the known values:

\[ \pi = \frac{(0.00511 \text{ mol}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (290.15 \text{ K})}{0.150 \text{ L}} \]

### Step 5: Calculate the osmotic pressure
Now, we perform the calculation:

\[ \pi = \frac{(0.00511) \times (0.0821) \times (290.15)}{0.150} \]

Calculating the numerator:

\[ 0.00511 \times 0.0821 \times 290.15 \approx 0.1233 \]

Now divide by the volume:

\[ \pi \approx \frac{0.1233}{0.150} \approx 0.822 \text{ atm} \]

### Final Answer
The osmotic pressure of the solution is approximately **0.82 atm**.

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