calculate the freezing point of a .05500 m aqueous solution of NaNO3
delta T = i*Kf*m
Substitute into the equation and solve for delta T. Then subtract from 0 C to find the freezing point.
i = 2 for NaNO3
Kf is 1.86 and you have m.
.2406
To calculate the freezing point of a solution, you can use the formula:
ΔTf = Kf * m
Where:
ΔTf = change in freezing point
Kf = cryoscopic constant (specific to the solvent)
m = molality of the solution
To find the freezing point, we first need to find the molality of the solution.
Molality (m) is defined as the number of moles of solute divided by the mass of the solvent (in kg).
To calculate the molality, we need to know the number of moles of NaNO3 and the mass of the solvent.
The molar mass of NaNO3 can be calculated as follows:
NaNO3: 22.99 g/mol (Na) + 14.01 g/mol (N) + (3 * 16.00 g/mol) (O)
NaNO3: 22.99 g/mol + 14.01 g/mol + 48.00 g/mol
NaNO3: 84.00 g/mol
Next, we calculate the number of moles of NaNO3:
moles of NaNO3 = concentration (m) * volume (L)
Given:
concentration = 0.05500 m
volume = 1 L (assuming 1 L of solution)
moles of NaNO3 = 0.05500 mol/L * 1 L
moles of NaNO3 = 0.05500 mol
Now, let's calculate the molality:
molality (m) = moles of solute / mass of solvent (in kg)
Given:
mass of solvent = mass of water
Assuming that the total mass of the solution is 1 kg (since water has a density close to 1 g/mL), the mass of water is also 1 kg.
molality = 0.05500 mol / 1 kg
molality = 0.05500 m
Lastly, we can calculate the change in freezing point:
ΔTf = Kf * m
The cryoscopic constant, Kf, for water is approximately 1.86 °C/molal.
ΔTf = 1.86 °C/molal * 0.05500 m
ΔTf = 0.1023 °C
The freezing point of the NaNO3 solution is reduced by 0.1023°C from the freezing point of pure water.
To calculate the freezing point of a solution, we can use the equation:
ΔTf = Kf * m
Where:
- ΔTf is the change in freezing point of the solvent
- Kf is the cryoscopic constant for the solvent
- m is the molality of the solution (moles of solute per kilogram of solvent)
To find the freezing point depression constant (Kf), we need to refer to a table or a reliable source for the given solvent, which in this case is water (H2O). The cryoscopic constant for water is approximately 1.86 °C/m.
Now, let's calculate the molality (m) of the solution:
- Given: concentration (molarity) = 0.05500 m (moles of solute per liter of solution)
- We need to convert the given molarity to molality (moles of solute per kilogram of solvent).
- The molecular weight of NaNO3 is:
- Sodium (Na): 22.99 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol (x3 since it appears three times in the formula)
- Total: 85.00 g/mol
- First, convert the concentration from moles per liter (M) to moles per kilogram (m):
- 0.05500 m NaNO3 x (1 L / 1000 mL) x (1000 mL / 1 kg) = 0.05500 moles of NaNO3 per kg of water
Now, we can plug the values into the formula to calculate the change in freezing point (ΔTf):
ΔTf = 1.86 °C/m * 0.05500 m
Calculating this, we find:
ΔTf = 0.1023 °C
Therefore, the freezing point of the solution is decreased by approximately 0.1023 °C. To find the actual freezing point, you would need to subtract this value from the freezing point of pure water, which is 0 °C. Thus, the freezing point of the aqueous NaNO3 solution is approximately -0.1023 °C.