what is the osmotic pressure produced by a 1.20M glucose (CH12O6) solution at 25 degree
Chris/Elsie/et al
pi = MRT
glucose C6H12O6
C6H12O6 = 2C2H5O6 + 2CO2
To calculate the osmotic pressure produced by a solution, you can use the equation for osmotic pressure:
π = i * M * R * T
Where:
π = osmotic pressure (in Pa or N/m^2)
i = van't Hoff factor - the number of particles formed per molecule in solution
M = molarity of the solution (in mol/L)
R = ideal gas constant = 0.0821 L * atm / (mol * K)
T = temperature (in Kelvin)
In this case, the solute is glucose (C6H12O6), which does not dissociate or form particles in solution, so the van't Hoff factor (i) is equal to 1.
Given:
M = 1.20 M (mol/L)
R = 0.0821 L * atm / (mol * K)
T = 25°C = 25 + 273.15 = 298.15 K
Substituting the values into the equation:
π = 1 * 1.20 M * 0.0821 L * atm / (mol * K) * 298.15 K
Calculating:
π ≈ 29.27 L * atm / mol
The osmotic pressure produced by a 1.20 M glucose solution at 25°C is approximately 29.27 L * atm / mol.
To calculate the osmotic pressure produced by a solution, you can use the formula:
π = i * M * R * T
Where:
π is the osmotic pressure
i is the van't Hoff factor (the number of particles formed when the solute dissociates)
M is the molarity of the solution (moles per liter)
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
Before we calculate the osmotic pressure, let's determine the van't Hoff factor for glucose. Glucose does not dissociate or ionize when dissolved in water, so it does not separate into multiple particles. Therefore, the van't Hoff factor (i) for glucose is 1.
Now, let's plug in the given values into the formula:
i = 1
M = 1.20 M
R = 0.0821 L·atm/mol·K
T = 25 + 273 = 298 K (temperature in Kelvin)
π = 1 * 1.20 * 0.0821 * 298
= 29.7434 atm (rounded to four decimal places)
Therefore, the osmotic pressure produced by a 1.20M glucose solution at 25 degrees Celsius is approximately 29.7434 atm.