Calculate the osmotic pressure of 0.5% solution of glucose (molecular mass 180) at `18^(@)C`. The value of solution constant is `0.0821 "litre-atm K"^(-1) "mol"^(-1)`

Answer

To calculate the osmotic pressure of a 0.5% solution of glucose at 18°C, we can follow these steps:

### Step 1: Convert the percentage concentration to molarity
A 0.5% (w/v) solution means there are 0.5 grams of glucose in 100 mL of solution.

To find the number of moles of glucose:
- Molecular mass of glucose = 180 g/mol
- Moles of glucose = mass (g) / molecular mass (g/mol) = 0.5 g / 180 g/mol = 0.00278 moles

Now, since the solution volume is 100 mL, we convert this to liters:
- Volume of solution = 100 mL = 0.1 L

Now, we can calculate the molarity (C):
- Molarity (C) = moles of solute / volume of solution (L) = 0.00278 moles / 0.1 L = 0.0278 M

### Step 2: Convert the temperature to Kelvin
The temperature in Celsius needs to be converted to Kelvin:
- T(K) = T(°C) + 273.15 = 18 + 273.15 = 291.15 K

### Step 3: Use the osmotic pressure formula
The formula for osmotic pressure (π) is given by:
\[ \pi = C \cdot R \cdot T \]
Where:
- \( C \) = molarity of the solution (0.0278 M)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (291.15 K)

### Step 4: Substitute the values into the formula
Now we can substitute the values into the osmotic pressure formula:
\[ \pi = 0.0278 \, \text{mol/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 291.15 \, \text{K} \]

### Step 5: Calculate the osmotic pressure
Calculating the above expression:
\[ \pi = 0.0278 \times 0.0821 \times 291.15 \]
\[ \pi \approx 0.671 \, \text{atm} \]

### Final Answer:
The osmotic pressure of the 0.5% glucose solution at 18°C is approximately **0.671 atm**.

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