At what temperature `(s)` a `5%` solution `(w//V)` of glucose will develop an osmotic pressure of `7 atm`?

Options

A. 33.94 K
B. 306.94 K
C. 273 K
D. `33.94^(@)C`

Answer

To solve the problem of determining the temperature at which a 5% (w/v) solution of glucose develops an osmotic pressure of 7 atm, we can follow these steps:

### Step 1: Understand the given information
We have:
- Osmotic pressure (π) = 7 atm
- Concentration of glucose solution = 5% w/v
- Molecular weight of glucose (C₆H₁₂O₆) = 180 g/mol

### Step 2: Convert the percentage concentration to a usable form
A 5% w/v solution means there are 5 grams of glucose in 100 mL of solution. Since we need to work in liters for the osmotic pressure formula, we convert 100 mL to liters:
- Volume (V) = 100 mL = 0.1 L

### Step 3: Use the osmotic pressure formula
The formula for osmotic pressure is given by:
\[ \pi = CRT \]
Where:
- π = osmotic pressure
- C = molarity (moles of solute per liter of solution)
- R = ideal gas constant (0.0821 L·atm/(K·mol))
- T = temperature in Kelvin

### Step 4: Calculate the molarity (C)
First, we need to calculate the number of moles of glucose in the solution:
\[ \text{Moles of glucose} = \frac{\text{mass}}{\text{molecular weight}} = \frac{5 \text{ g}}{180 \text{ g/mol}} = \frac{5}{180} \text{ mol} \]

Now, we can find the molarity (C):
\[ C = \frac{\text{moles of glucose}}{\text{volume in liters}} = \frac{5/180}{0.1} = \frac{5}{18} \text{ mol/L} \]

### Step 5: Substitute values into the osmotic pressure equation
Now we can substitute the values into the osmotic pressure equation:
\[ 7 \text{ atm} = \left(\frac{5}{18} \text{ mol/L}\right) \cdot (0.0821 \text{ L·atm/(K·mol)}) \cdot T \]

### Step 6: Solve for T
Rearranging the equation to solve for T:
\[ T = \frac{7 \text{ atm}}{\left(\frac{5}{18}\right) \cdot (0.0821)} \]

Calculating the right side:
\[ T = \frac{7}{\left(\frac{5}{18}\right) \cdot (0.0821)} = \frac{7 \cdot 18}{5 \cdot 0.0821} \]

Calculating this gives:
\[ T \approx 306.94 \text{ K} \]

### Step 7: Convert to Celsius if needed
To convert Kelvin to Celsius:
\[ T(°C) = T(K) - 273.15 \approx 306.94 - 273.15 \approx 33.79 °C \]

### Final Answer
The temperature at which a 5% solution of glucose will develop an osmotic pressure of 7 atm is approximately **306.94 K** or **33.79 °C**.