The freezing point of an aqueous solution of glucose was measured to be -1.52 oC. What is the osmotic pressure, in atm, of 100 mL of this solution at 25.0 oC? (Kf = 1.86 oC/m)
When I calculated it out I got 16.79 atm but the answer was wrong don't know what I messed up on.
Well, when it comes to calculating osmotic pressure, you know what they say... It's all about the pressure, 'bout the pressure, no trouble!
Now, let's dive in and figure out what went wrong with your calculation. You started off right by finding the temperature difference: ΔT = T_f - T_i = (-1.52 oC - 0 oC) = -1.52 oC.
Now, here comes the fun part: we need to convert this temperature difference into molality, using the cryoscopic constant (K_f). Remember, it's all about constant conversions 'round here!
ΔT = K_f*m
-1.52 oC = 1.86 oC/m * m
Dividing both sides by 1.86 oC/m, we find:
m = -1.52 oC / 1.86 oC/m
m ≈ -0.817 m
Now, we have the molality, but we need to convert it into moles of solute per kilogram of solvent to calculate the osmotic pressure.
molality (m) = moles of solute / kg of solvent
Since we have 100 mL of solution, we can convert it to kg:
100 mL = 0.1 L = 0.1 kg (assuming density = 1 g/mL).
So, moles of solute = molality * kg of solvent = -0.817 m * 0.1 kg = -0.0817 moles.
Now, let's calculate the osmotic pressure using the ideal gas law:
π = (n/V) * (RT)
π = (-0.0817 moles / 0.1 L) * (0.0821 L·atm/mol·K) * (298 K)
Calculating this gives us:
π ≈ -2.42 atm
Ah, looks like we forgot to take into account that the molality should always be positive! After considering the absolute value, we get the correct osmotic pressure:
π ≈ 2.42 atm.
So, my friend, you were so close! Keep your chin up and remember: in the world of chemistry, a little laughter can help you get through any molar problem!
To calculate the osmotic pressure (π) of the solution, we can use the equation:
π = MRT
Where:
M = molarity of the solution (in mol/L)
R = ideal gas constant (0.0821 L · atm/(mol · K))
T = temperature in Kelvin (25.0 oC = 25 + 273 = 298 K)
First, let's determine the molarity of the solution. We can use the formula:
ΔT = Kf · m
Rearranging the equation, we have:
m = ΔT / Kf
Where:
ΔT = change in temperature = -1.52 oC
Kf = cryoscopic constant = 1.86 oC/m
Substituting the values, we get:
m = -1.52 oC / 1.86 oC/m
m ≈ -0.82 m
Since molarity cannot be negative, the negative sign indicates that the solution is dilute. Therefore, we will consider the magnitude of m, which is 0.82 m.
Now, let's substitute the values into the osmotic pressure equation:
π = MRT
π = (0.82 mol/L) × (0.0821 L · atm/(mol · K)) × (298 K)
π ≈ 19.86 atm
The osmotic pressure of the solution is approximately 19.86 atm.
It appears that you made an error while calculating the molarity of the solution (m). Check your calculations for ΔT / Kf again and make sure to consider the magnitude of the value.
To calculate the osmotic pressure of the solution, you need to use the van't Hoff equation. The equation relates the osmotic pressure (π) to the concentration of solute particles (in this case glucose), the gas constant (R), and the temperature (T) in Kelvin.
The van't Hoff equation can be written as:
π = i * M * R * T
where:
π = osmotic pressure
i = van't Hoff factor (the number of particles per formula unit in the solution)
M = molarity of the solution
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, let's calculate the van't Hoff factor.
Glucose (C6H12O6) is a non-electrolyte, so it does not dissociate into ions in solution. Therefore, the van't Hoff factor (i) for glucose is 1.
Next, convert the temperature to Kelvin:
25.0 oC + 273.15 = 298.15 K
Now, we need to determine the molarity of the solution (M). We can use the freezing point depression equation to find this value:
ΔT = Kf * m
where:
ΔT = change in temperature (freezing point depression)
Kf = freezing point depression constant
m = molality of the solution (moles of solute per kilogram of solvent)
First, calculate the change in temperature:
ΔT = (freezing point of solvent - freezing point of solution)
= (0 oC - (-1.52 oC))
= 1.52 oC
Now, plug in the values into the equation to solve for m:
1.52 oC = (1.86 oC/m) * m
(molality = moles of solute / kg of solvent)
Solving for m:
m = 1.52 oC / 1.86 oC/m
m = 0.817 mol/kg
Since the volume of the solution is given as 100 mL, you need to convert it to kilograms:
100 mL = 0.1 L
Assuming the density of the solution is 1 g/mL, the mass of the solution would be 100 g or 0.1 kg.
Finally, calculate the osmotic pressure using the van't Hoff equation:
π = i * M * R * T
= 1 * (0.817 mol/kg) * (0.0821 L·atm/(mol·K)) * 298.15 K
= 20.05 atm
Therefore, the osmotic pressure of the solution is 20.05 atm, not 16.79 atm. It seems like there was a calculation error in your previous attempt.