a solution is prepared by dissolving 4.9 g of sucrose (c12h22011) in 175 g of water. calculate the boiling point and osmotic pressure of this solution at 25°c. assume that molarity and molality of this solution are equal. e(h,0) = 0.52.
See your other post above.
Well, well, well! Looks like we have a scientific question here. Don't worry, I'll do my best to put a funny twist on it!
To calculate the boiling point and osmotic pressure of this solution, we first need to find the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.
First, let's figure out the moles of sucrose we have. Since the molar mass of sucrose (C12H22011) is 342.3 g/mol, we can divide 4.9 g by 342.3 g/mol to get the moles, which is approximately 0.014 moles.
Now, let's calculate the molality. We divide the number of moles of sucrose by the mass of water in kilograms. Since 1 g = 0.001 kg, the mass of water is 175 g * 0.001 kg/g = 0.175 kg. So, the molality is 0.014 moles / 0.175 kg = 0.08 mol/kg.
Next, let's calculate the boiling point elevation using the equation: ΔTb = Kbm. ΔTb is the boiling point elevation, Kb is the molal boiling point constant, and m is the molality.
Since you didn't provide the value of the molal boiling point constant (Kb), I won't be able to give you a specific answer here. But rest assured, you can find the value in a reference book or on the internet.
As for the osmotic pressure, we can use the equation: π = iMRT. Here, π represents the osmotic pressure, 'i' is the van't Hoff factor (which is the number of particles the solute dissociates into), M is the molar concentration, R is the ideal gas constant, and T is the temperature in kelvin.
Again, since you didn't provide the van't Hoff factor or the molar concentration (molarity), I can't calculate the osmotic pressure for you. But remember, humor is the osmotic pressure relief valve in life!
I hope this at least gave you a smile while you search for the values you need. And if you have any other questions, I'll be here, ready to clown around!
To calculate the boiling point of the solution, we can use the equation:
ΔTb = Kb * m
where ΔTb is the boiling point elevation, Kb is the molal boiling point constant for water (0.512 °C/m), and m is the molality of the solution.
First, we need to find the molality of the solution using the given data:
Molar mass of sucrose (C12H22O11) = 342.30 g/mol
Number of moles of sucrose = 4.9 g / 342.30 g/mol = 0.014 mol
Mass of water = 175 g
Molality (m) = moles of solute / mass of solvent (in kg)
m = 0.014 mol / 0.175 kg = 0.08 mol/kg
Now we can calculate the boiling point elevation:
ΔTb = 0.512 °C/m * 0.08 mol/kg = 0.041 °C
Therefore, the boiling point of the solution is increased by 0.041 °C.
To calculate the osmotic pressure of the solution, we can use the equation:
π = MRT
where π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant (0.0821 L⋅atm/(mol⋅K)), and T is the temperature in Kelvin.
Since the molarity and molality of this solution are equal, we can use the same value, which is equal to the molality calculated earlier.
Molarity (M) = molality (m) = 0.08 mol/L
Temperature (T) = 25 °C = 298 K
Now we can calculate the osmotic pressure:
π = 0.08 mol/L * 0.0821 L⋅atm/(mol⋅K) * 298 K = 1.95 atm
Therefore, the osmotic pressure of this solution at 25 °C is 1.95 atm.
To calculate the boiling point and osmotic pressure of the solution, we need to use the formulas for these properties. Let's go step by step.
1. Boiling point elevation (ΔTb):
The boiling point elevation is given by the equation: ΔTb = Kb * m * i, where ΔTb is the boiling point elevation, Kb is the molal boiling point constant, m is the molality of the solution, and i is the van't Hoff factor.
First, we need to calculate the molality of the solution (m).
Molality (m) = moles of solute / mass of solvent (in kg)
To find the moles of solute (sucrose), we use the molar mass of sucrose (C12H22O11):
molar mass of sucrose = (12 * 12.01 g/mol) + (22 * 1.01 g/mol) + (11 * 16.00 g/mol)
= 342.34 g/mol
moles of sucrose = mass of sucrose / molar mass of sucrose
= 4.9 g / 342.34 g/mol
Now, we need to convert the mass of water to kg:
mass of water = 175 g = 0.175 kg
Now we can calculate the molality (m):
m = moles of sucrose / mass of water
= (4.9 g / 342.34 g/mol) / 0.175 kg
2. Van't Hoff factor (i):
The van't Hoff factor (i) is a measure of the number of particles the solute dissociates into in a solution. For sucrose, it does not dissociate, so the van't Hoff factor for sucrose is 1.
3. Molal boiling point constant (Kb):
Kb varies depending on the solvent, in this case, water. For water, Kb = 0.512 °C/m.
Now we can calculate the boiling point elevation (ΔTb):
ΔTb = Kb * m * i
ΔTb = 0.512 °C/m * (4.9 g / 342.34 g/mol) / 0.175 kg * 1
4. Boiling point (Tb):
The boiling point of the solution can be found by adding the boiling point elevation (ΔTb) to the boiling point of the pure solvent (water). The boiling point of pure water at 1 atm is 100°C.
Tb = 100°C + ΔTb
Now, to calculate the osmotic pressure:
5. Osmotic pressure (π):
Osmotic pressure is given by the equation: π = i * M * R * T, where π is the osmotic pressure, i is the van't Hoff factor, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the problem states that molarity and molality are equal, we can use the calculated value of molality for the molarity.
To convert molality to molarity, we need to consider the density of water and its molar mass:
density of water at 25°C = 0.997 g/ml = 0.997 g/cm^3
molar mass of water = (2 * 1.01 g/mol) + (16.00 g/mol)
= 18.02 g/mol
molarity (M) = molality (m) * density of water / molar mass of water
Now we can calculate the osmotic pressure (π):
π = i * M * R * T
= 1 * (4.9 g / 342.34 g/mol) * density of water / molar mass of water * R * (25 + 273.15) K
Where R is the ideal gas constant, R = 0.0821 L.atm/mol.K.
Plug in the values and calculate the boiling point and osmotic pressure.