What is the boiling point of an aqueous solution of a non electrolyte that has an osmotic pressure of 10.50atm at 25 degrees C? Kb of water is 0.52 degrees C/m. Assume the density of the solution is the same as that of pure water.
pi=osmotic pressure = Molarity*R*T
Plug in the numbers and calculate molarity.
Then use delta T = Kb*m
since the density of the solution is 1 kg/L, you may use m = M and solve for delta T.
What is the boiling point of a 4.00 m aqueous solution of a nonvolatile nonelectrolyte? (The boiling point elevation constant for water is 0.512° C/m).
To find the boiling point of an aqueous solution of a non-electrolyte, we can use the relationship between osmotic pressure and boiling point elevation.
The equation we can use is ΔTb = i * Kb * m, where:
- ΔTb is the boiling point elevation (difference between the boiling point of the solution and pure solvent),
- i is the van't Hoff factor (the number of particles formed in solution from each molecule of the solute, which is 1 for non-electrolytes),
- Kb is the molal boiling point elevation constant (given as 0.52 degrees C/m for water),
- m is the molality of the solution (moles of solute per kilogram of solvent).
We have the osmotic pressure, which is related to the molarity of the solution (M) by the equation π = M * R * T, where:
- π is the osmotic pressure,
- R is the ideal gas constant (0.0821 L * atm / mol * K),
- T is the temperature in Kelvin (25 degrees Celsius = 298 K).
We need to convert the osmotic pressure to molarity and then to molality to use in the boiling point elevation equation.
Step 1: Calculate the molarity (M) from the osmotic pressure:
π = M * R * T
10.50 atm = M * (0.0821 L * atm / mol * K) * 298 K
M = 10.50 atm / (0.0821 L * atm / mol * K * 298 K)
M ≈ 0.430 mol/L
Step 2: Convert molarity (M) to molality (m):
Molality (m) = M / (density of water in g/mL)
Since the density of the solution is the same as that of pure water, the density is 1 g/mL.
m = 0.430 mol/L / 1 g/mL
m ≈ 0.430 mol/kg
Step 3: Calculate the boiling point elevation (ΔTb) using the molality (m) and the boiling point elevation constant (Kb):
ΔTb = i * Kb * m
ΔTb = (1) * (0.52 degrees C/m) * (0.430 mol/kg)
ΔTb ≈ 0.224 degrees C
Step 4: Calculate the boiling point of the solution:
Boiling point of solution = boiling point of pure solvent + ΔTb
Since the boiling point of pure water is 100 degrees Celsius, the boiling point of the solution is:
Boiling point of solution = 100 degrees C + 0.224 degrees C
Boiling point of solution ≈ 100.224 degrees C
Therefore, the boiling point of the aqueous solution is approximately 100.224 degrees Celsius.
To find the boiling point of an aqueous solution, you can use the formula:
ΔTb = Kbm
where:
ΔTb is the boiling point elevation
Kb is the boiling point elevation constant (molar boiling point constant)
m is the molality of the solution (moles of solute per kg of solvent)
In this case, the osmotic pressure (π) of the solution is given as 10.50 atm. The osmotic pressure is related to the molarity of the solution (M) by the equation:
π = MRT
where R is the gas constant and T is the temperature in Kelvin.
Since we are given the osmotic pressure at 25 degrees Celsius, we need to convert it to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 + 273.15
T(K) = 298.15 K
We also need to find the molality (m) of the solution. Molality is defined as the moles of solute (in this case, non-electrolyte) per kg of solvent. Since the density of the solution is assumed to be the same as that of pure water, the mass of the solvent is equal to the mass of water.
To find the molality, we need to know the moles of solute. Unfortunately, that information is not provided in the question. Hence, we cannot determine the boiling point of the solution without additional information.
If you have the moles of solute, you can calculate the molality as follows:
molality (m) = moles of solute / mass of solvent (in kg)
Once you have the molality, you can calculate the boiling point elevation (ΔTb) using the formula mentioned earlier.