calculate the osmotic pressure of a solution containing 01g/mol KCL/100g water @ 25C⁰given the density of a water @ 25C⁰ is 0.997kg/m³

To calculate the osmotic pressure of a solution, we can use the formula derived from van 't Hoff's law, which states that osmotic pressure (π) is directly proportional to the molarity (C) of the solute in the solution. The formula is given by:

π = CRT

Where:

• π = osmotic pressure (in atm)
• C = molarity of the solution (in mol/L)
• R = universal gas constant (0.0821 L·atm/(K·mol))
• T = temperature in Kelvin (K)

Let's break down the calculation step by step:

Step 1: Convert the temperature to Kelvin

The temperature given is 25°C. To convert this to Kelvin, we add 273.15:

T = 25 + 273.15 = 298.15 K

Step 2: Calculate the number of moles of KCl

The molar mass of KCl (potassium chloride) is approximately 74.55 g/mol. Given that we have 0.1 g of KCl, we can find the number of moles:

Number of moles = mass (g) / molar mass (g/mol)

Number of moles = 0.1 g / 74.55 g/mol ≈ 0.00134 mol

Step 3: Calculate the volume of the solution

We have 100 g of water, and we need to convert this to liters. Given the density of water at 25°C is 0.997 kg/m³, we can find the volume:

Volume (L) = mass (kg) / density (kg/m³)

Volume = 0.1 kg / 0.997 kg/m³ ≈ 0.1003 L

Step 4: Calculate the molarity of the KCl solution

Molarity (C) is defined as the number of moles of solute per liter of solution:

C = moles of solute / volume of solution (L)

C = 0.00134 mol / 0.1003 L ≈ 0.01335 mol/L

Step 5: Calculate the osmotic pressure

Now we can substitute the values into the osmotic pressure formula:

π = CRT

π = (0.01335 mol/L) × (0.0821 L·atm/(K·mol)) × (298.15 K)

π ≈ 0.328 atm

Thus, the osmotic pressure of the KCl solution at 25°C is approximately 0.328 atm. This value indicates how much pressure would be required to stop the flow of solvent into the solution through a semipermeable membrane, which is a key concept in understanding osmotic processes in biological systems and various chemical applications.