What is the osmotic pressure in torr of a 0.0155 M glucose solution at body temperature ( 35.7 °C)? What are the boiling and freezing points for the same solution?
pi = MRT
Solve for pi (osmotic pressure) in atm and convert to torr. 760 torr = 1 atm.
If we assume the m = M, then
delta T = Kf*m, solve for dT and subtract from zero to obtain the new freezing point. This won't be exact since m is not quite equal to M but it will be close. There is no density listed; therefore, the assumption must be made.
delta T = Kb*m. Solve for dT and add to 100 to obtain the new boiling point.
To calculate the osmotic pressure, boiling point, and freezing point of a solution, we need to use the formulas and concepts related to colligative properties.
For osmotic pressure (𝜋), we can use the equation:
𝜋 = (n/V)RT
where:
- 𝜋 is the osmotic pressure
- n is the moles of solute
- V is the volume of the solution in liters
- R is the ideal gas constant (0.0821 L.atm/mol.K)
- T is the absolute temperature in Kelvin
To calculate the boiling point elevation (∆Tb), we can use the equation:
∆Tb = Kbm
where:
- ∆Tb is the change in boiling point
- Kb is the molal boiling point constant
- m is the molality of the solution in moles solute/kg solvent
To calculate the freezing point depression (∆Tf), we can use the equation:
∆Tf = Kfm
where:
- ∆Tf is the change in freezing point
- Kf is the molal freezing point constant
- m is the molality of the solution in moles solute/kg solvent
Now, let's calculate the osmotic pressure, boiling point elevation, and freezing point depression for a 0.0155 M glucose solution at 35.7 °C (which needs to be converted to Kelvin).
Firstly, let's convert the temperature to Kelvin:
35.7 °C + 273.15 = 308.85 K
Now, let's calculate the osmotic pressure:
𝜋 = (n/V)RT
Since we are not given the volume of the solution, we cannot calculate the osmotic pressure.
Next, let's calculate the boiling and freezing points using the formulas provided, but we need to know the molal boiling point constant (Kb) and molal freezing point constant (Kf) for glucose in order to proceed.
Without those constants, we won't be able to determine the specific values for the boiling and freezing points of the glucose solution.