When 24.0g og glucose(a nonelectrolyte) are dissolved in 500g of water, the solution has a freezing point of -0.47C. What is the molar mass of glucose according to yhe data?Kf of water is 1.86C/m.
delta T = Kf*molality
Substitute and solve for molality.
m = moles/kg solvent
Substitute and solve for moles.
moles = grams/molar mass
Substitute and solve for molar mass.
To find the molar mass of glucose, we can use the following formula:
Molar mass of solute (glucose) = (ΔTf * m) / (Kf * moles of solute)
First, let's calculate the change in freezing point (ΔTf):
ΔTf = freezing point of pure solvent - freezing point of solution
The freezing point of pure water is 0°C, and the freezing point of the solution is -0.47°C. Therefore, ΔTf = 0 - (-0.47) = 0.47°C.
Next, we calculate the molality (m) of the solution:
molality (m) = moles of solute / mass of solvent in kg
The mass of the solvent (water) is given as 500 g, which is equivalent to 0.5 kg. Since glucose is a nonelectrolyte, it does not dissociate, so the number of moles of solute is equal to the number of moles of glucose.
Using the formula for molality, we can rearrange it to find the moles of solute:
moles of solute = m * mass of solvent in kg
We can substitute the given values into the equation: moles of solute = m * 0.5 kg. However, we don't have the value of molality, so let's calculate it using the formula:
molality (m) = moles of solute / mass of solvent in kg
Since we have 24.0 g of glucose, we convert it to kg by dividing by 1000: mass of solute = 24.0 g / 1000 = 0.024 kg.
Now we can calculate the molality:
molality (m) = 0.024 kg / 0.5 kg = 0.048 mol/kg
Now we can substitute all the known values into the molar mass formula:
Molar mass of glucose = (ΔTf * m) / (Kf * moles of solute)
Molar mass of glucose = (0.47°C * 0.048 mol/kg) / (1.86°C/m * 0.024 kg)
Simplifying the equation:
Molar mass of glucose = (0.02376 mol °C / kg) / (0.04464 mol °C / kg)
Now we can divide the values:
Molar mass of glucose = 0.531 kg/mol
Therefore, the molar mass of glucose according to the given data is approximately 0.531 kg/mol.
To find the molar mass of glucose, we need to use the equation for freezing point depression:
ΔT = Kf * m * i,
where:
ΔT is the change in freezing point of the solution,
Kf is the molal freezing point depression constant of the solvent (water),
m is the molality of the solution, and
i is the van't Hoff factor, which represents the number of particles formed in the solution.
In this case, glucose is a nonelectrolyte, so i = 1.
Given:
ΔT = -0.47°C,
Kf = 1.86°C/m, and
mass of solvent (water) = 500 g.
First, we need to calculate the molality (m) using the formula:
m = moles of solute / mass of solvent (in kg).
Mass of glucose = 24.0 g.
Molar mass of glucose (M) = ?
Number of moles of glucose = mass / molar mass:
moles of glucose = 24.0 g / M.
Now, we can calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
m = (24.0 g / M) / (500 g / 1000) (converting mass of solvent to kg)
m = (24.0 g / M) / 0.5 kg
m = 48.0 g / (M kg)
Now, substitute the values into the freezing point depression equation to find the molar mass (M) of glucose:
ΔT = Kf * m * i
-0.47°C = 1.86°C/m * (48.0 g / M kg) * 1
-0.47°C = 1.86°C/m * (48.0 / M)
-0.47°C = 89.28 / M
Now, let's solve for M:
-0.47°C = 89.28 / M
-0.47°C * M = 89.28
-0.47M = 89.28
M = 89.28 / -0.47
M ≈ 190.09 g/mol
Therefore, the molar mass of glucose is approximately 190.09 g/mol according to the given data.