1)determine the molality for an antifreeze solutions. Kf for water is 1.86 C/m


in the experiment we found the freezing point of the antifreeze it was 1.2 C

2) Determine the molar mass. 5.00g of antifreeze/50g of water

3)calculate percent error.

1) To determine the molality of the antifreeze solution, we will use the formula:

molality (m) = ΔTf / Kf

Where:
ΔTf = change in freezing point = Tf - Tf0
Kf = molal freezing point depression constant

Given:
Kf = 1.86 °C/m
Tf = 1.2 °C (freezing point of the antifreeze)

Calculating the ΔTf:
ΔTf = Tf - Tf0
ΔTf = 1.2 °C - 0 °C (since pure water freezes at 0 °C)
ΔTf = 1.2 °C

Now, substitute the values into the formula:

molality (m) = (1.2 °C) / (1.86 °C/m)

The molality of the antifreeze solution is calculated to be approximately 0.645 m.

2) To determine the molar mass, we can use the formula:

molar mass (M) = (mass of solute in grams) / (moles of solute)

Given:
Mass of antifreeze = 5.00 g
Mass of water = 50 g

Since we have the mass of antifreeze in grams, we need to find the number of moles of antifreeze.

Using the formula:

moles of solute = (mass of solute in grams) / (molar mass of solute)

We can rearrange the formula to solve for the molar mass:

molar mass of solute = (mass of solute in grams) / (moles of solute)

Finding the moles of antifreeze:
moles of antifreeze = (mass of antifreeze) / (molar mass of antifreeze)

Substituting the values:
moles of antifreeze = 5.00 g / (molar mass of antifreeze)

Now, we know that the total mass of the solution is the sum of the mass of the antifreeze and water:

Total mass = Mass of antifreeze + Mass of water
Total mass = 5.00 g + 50 g
Total mass = 55.00 g

Since we have the total mass in grams, we can find the number of moles of water:

moles of water = (mass of water) / (molar mass of water)

Substituting the values:
moles of water = 50 g / (molar mass of water)

Since water is the solvent, we can assume its molar mass is 18.015 g/mol.

Now, we can write the expression for the molar mass:

molar mass = (5.00 g / (molar mass of antifreeze)) + (50 g / 18.015 g/mol)

Note: You will need to substitute the specific molar mass of the antifreeze into the equation to calculate the molar mass.

3) To calculate the percentage error, we need to compare the experimental value with the accepted or theoretical value.

The theoretical value of the molality (m) can be calculated based on the molar mass of the antifreeze using the equation from step 2:

molality (m) = (mass of antifreeze) / (moles of antifreeze)

The experimental molality is given in step 1 as 0.645 m.

To calculate the percent error, we use the formula:

% error = | (experimental value - theoretical value) / theoretical value | * 100

Substitute the values into the equation to calculate the percent error.

1) To determine the molality of the antifreeze solution, you need to use the formula:

Molality (m) = ΔTf / Kf

where:
ΔTf is the change in freezing point (in degrees Celsius)
Kf is the freezing point depression constant (in degrees Celsius per molality, or C/m)

In this case, the freezing point depression constant (Kf) for water is given as 1.86 C/m, and the change in freezing point (ΔTf) is 1.2 C.

Plugging in the values into the formula, we get:

Molality (m) = 1.2 C / (1.86 C/m)

Molality (m) ≈ 0.645 m

Therefore, the molality of the antifreeze solution is approximately 0.645 m.

2) To determine the molar mass of the antifreeze, you need to use the formula:

Molar mass (M) = (mass of solute / moles of solute) = (mass of solute / molality)

In this case, the mass of the antifreeze solute is given as 5.00 g, and the molality of the solution is 0.645 m.

Plugging in the values into the formula, we get:

Molar mass (M) = 5.00 g / 0.645 m

Molar mass (M) ≈ 7.752 g/mol

Therefore, the molar mass of the antifreeze is approximately 7.752 g/mol.

3) To calculate the percent error, you need to compare the experimental value with the accepted value and calculate the difference.

In this case, let's assume the accepted value for the molar mass of the antifreeze is 10.00 g/mol (just as an example).

The formula to calculate percent error is:

Percent error = | (experimental value - accepted value) / accepted value | x 100%

Plugging in the values, we get:

Percent error = | (7.752 g/mol - 10.00 g/mol) / 10.00 g/mol | x 100%

Percent error ≈ 22.48%

Therefore, the percent error between the experimental molar mass and the accepted molar mass is approximately 22.48%.