Calculate the osmotic pressure in atm of the solution 0.1M NaCl at 25 degree celcius. Assume a 100% ionization solute

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To calculate the osmotic pressure of a solution, you can use the formula:

π = MRT

Where:
π = osmotic pressure (in atm)
M = molarity of the solute (in mol/L)
R = ideal gas constant (value is 0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

Given:
Molarity (M) = 0.1M
Temperature (T) = 25°C = 25 + 273 = 298 K
Ideal gas constant (R) = 0.0821 L·atm/(mol·K)

Substituting the values into the formula:

π = (0.1)(0.0821)(298)
π = 2.433 atm

Therefore, the osmotic pressure of the 0.1M NaCl solution at 25°C is approximately 2.433 atm.

To calculate the osmotic pressure, you can use the formula:

Osmotic Pressure (π) = Molarity (M) * Gas Constant (R) * Temperature (T)

First, let's convert the temperature from Celsius to Kelvin:

T (Kelvin) = T (Celsius) + 273.15
T = 25 + 273.15 = 298.15 K

Next, we need to calculate the molarity of the NaCl solution:

Molarity (M) = 0.1 M

The gas constant (R) has a value of 0.0821 L·atm/mol·K.

Now, we can plug in the values into the formula:

π = 0.1 M * 0.0821 L·atm/mol·K * 298.15 K

Calculating this, we get:

π = 2.467 atm

Therefore, the osmotic pressure of a 0.1M NaCl solution at 25 degrees Celsius is approximately 2.467 atm.