Calculate the osmotic pressure of 0.2 M glucose solution at 300 K.
`(R=8.314 J mol^(-1) K^(-1))`

Explanation

To calculate the osmotic pressure of a 0.2 M glucose solution at 300 K, we will use the formula for osmotic pressure, which is given by:

\[ \pi = CRT \]

where:
- \(\pi\) = osmotic pressure
- \(C\) = concentration of the solution in moles per cubic meter (mol/m³)
- \(R\) = ideal gas constant (8.314 J mol⁻¹ K⁻¹)
- \(T\) = temperature in Kelvin (K)

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Concentration of glucose solution, \(C = 0.2 \, \text{M}\)
- Temperature, \(T = 300 \, \text{K}\)
- Gas constant, \(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\)

**Step 2: Convert the concentration from molarity to moles per cubic meter.**
- Molarity (M) is defined as moles per liter. Since \(1 \, \text{M} = 1 \, \text{mol/L}\), we convert it to moles per cubic meter:
\[ C = 0.2 \, \text{mol/L} = 0.2 \, \text{mol/dm}^3 = 0.2 \times 10^3 \, \text{mol/m}^3 = 200 \, \text{mol/m}^3 \]

**Step 3: Substitute the values into the osmotic pressure formula.**
\[ \pi = CRT = (200 \, \text{mol/m}^3) \times (8.314 \, \text{J mol}^{-1} \text{K}^{-1}) \times (300 \, \text{K}) \]

**Step 4: Calculate the osmotic pressure.**
\[ \pi = 200 \times 8.314 \times 300 \]
\[ \pi = 499,800 \, \text{Pa} \quad \text{(since 1 J/m³ = 1 Pa)} \]
\[ \pi \approx 4.998 \times 10^5 \, \text{Pa} \]

**Step 5: Final result.**
The osmotic pressure of the 0.2 M glucose solution at 300 K is approximately:
\[ \pi \approx 4.998 \times 10^5 \, \text{Pa} \quad \text{or} \quad 4.998 \times 10^5 \, \text{N/m}^2 \]