At `17^(@)C` the osmotic pressure of an aqueous of sucrose is `1.855 atm` per `150 mL` solution. Calculate the weight of sucrose in solution.

Explanation

To solve the problem of calculating the weight of sucrose in a solution with a given osmotic pressure, we can follow these steps:

### Step 1: Understand the Formula for Osmotic Pressure
The osmotic pressure (\( \pi \)) of a solution is given by the formula:
\[ \pi = \frac{nRT}{V} \]
where:
- \( \pi \) = osmotic pressure (in atm)
- \( n \) = number of moles of solute
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin
- \( V \) = volume of the solution in liters

### Step 2: Rearranging the Formula
We can rearrange the formula to find the number of moles (\( n \)):
\[ n = \frac{\pi V}{RT} \]

### Step 3: Convert Volume to Liters
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 4: Convert Temperature to Kelvin
The temperature is given as 17°C. To convert this to Kelvin:
\[ T = 17 + 273 = 290 \text{ K} \]

### Step 5: Substitute Values into the Formula
Now we can substitute the values into the rearranged formula:
\[ n = \frac{1.855 \text{ atm} \times 0.150 \text{ L}}{0.0821 \text{ L·atm/(K·mol)} \times 290 \text{ K}} \]

### Step 6: Calculate the Number of Moles
Calculating the above expression:
\[ n = \frac{1.855 \times 0.150}{0.0821 \times 290} \]
\[ n = \frac{0.27825}{23.849} \approx 0.01165 \text{ moles} \]

### Step 7: Calculate the Weight of Sucrose
The weight of sucrose can be calculated using the formula:
\[ \text{Weight} = n \times \text{Molar Mass} \]
The molar mass of sucrose (C₁₂H₂₂O₁₁) is approximately 342 g/mol. Therefore:
\[ \text{Weight} = 0.01165 \text{ moles} \times 342 \text{ g/mol} \approx 3.99 \text{ g} \]

### Final Answer
The weight of sucrose in the solution is approximately **4 grams**.

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