A tube of uniform cross-sectional area `1 cm^(2)` is closed at one end with semi-permeable membrane. A solution of `5 g` glucose per `100 mL` is placed inside the tube and is dipped in pure water at `27^(@)C`. When equilibrium is established, calculate:
a. The osmotic pressure of solution.
b.The height developed in vertical column.
Assume the density of final glucose solution `1 g mL^(-1)`

Explanation

To solve the problem step-by-step, we will first calculate the osmotic pressure of the glucose solution and then determine the height of the vertical column developed due to this osmotic pressure.

### Step 1: Calculate the Osmotic Pressure (π)

**Given:**
- Mass of glucose (W) = 5 g
- Volume of solution (V) = 100 mL = 0.1 L
- Temperature (T) = 27°C = 300 K (273 + 27)
...