Calculate the mole fraction of fructose (molar mass = 180.2g/mol) in an aqueous solution that has a freezing point of -7.0 degrees Celsius.
freezing pt depression=1.86*m
m= 7/1.86=3.76
so in a kg of water, mole fraction=3.76/(1000/18+3.76) note, the denominator is total moles. 6.3 percent
To calculate the mole fraction of fructose in an aqueous solution, we need to first determine the freezing point depression caused by the presence of fructose in the solution. This can be done using the formula:
ΔTf = Kf * m
where:
ΔTf = freezing point depression
Kf = cryoscopic constant of the solvent
m = molality of the solute
For water as the solvent, the cryoscopic constant (Kf) is approximately 1.86 °C/m.
Given that the freezing point depression (ΔTf) is 7.0 °C and water as the solvent, we can calculate the molality (m) of the fructose using the equation:
m = ΔTf / Kf
m = 7.0 °C / 1.86 °C/m
m ≈ 3.76 mol/kg
Next, we can calculate the moles of fructose (n) present in the solution. We know the molar mass of fructose is 180.2 g/mol.
n = m * kg of solvent
Let's assume we have 1 kg of water as the solvent (since we were given an aqueous solution), then:
n = 3.76 mol/kg * 1 kg
n = 3.76 mol
Finally, we can calculate the mole fraction (X) of fructose using the equation:
X = n / (n water + n fructose)
Since we assumed we have 1 kg of water, the moles of water (n water) can be calculated as:
n water = 1 kg * (1000 g/kg) / (18.015 g/mol)
n water ≈ 55.495 mol
Now, we can calculate the mole fraction of fructose:
X = 3.76 mol / (55.495 mol + 3.76 mol)
X ≈ 0.063
Therefore, the mole fraction of fructose in the aqueous solution is approximately 0.063.
To calculate the mole fraction of fructose in an aqueous solution, you need to know the freezing point depression constant (Kf) of the solvent and the change in freezing point (ΔTf) caused by the solute.
The freezing point depression (ΔTf) is given by the formula:
ΔTf = Kf * m
where Kf is the freezing point depression constant of the solvent, and m is the molality of the solute.
The molality (m) is the amount of solute (in moles) divided by the mass of the solvent (in kilograms). It is given by the formula:
m = n / (m_solute * kg_solvent)
To calculate the moles of fructose, you need to know the mass of fructose (in grams), and divide it by the molar mass (in grams per mole):
n = mass_fructose / molar_mass
Given that the freezing point depression (ΔTf) is -7.0 degrees Celsius, and the freezing point depression constant (Kf) for water is 1.86 °C/m, we can now calculate the molality (m) using the formula:
ΔTf = Kf * m -> -7.0 °C = 1.86 °C/m * m
Solving for m, we find:
m = -7.0 °C / 1.86 °C/m = -3.76 m
Since the molality of the solute is a negative value, we can conclude that no fructose is present in the solution, or the information provided is incorrect.
Therefore, the mole fraction of fructose in the aqueous solution is 0.