# Five (5.00) grams of glucose, C6H12)6, is dissolved in 500.0 grams of acetic acid. What is the new freezing point and boiling point for the solution?

Kf acetic acid = 3.90, Kb acetic acid = 3.07

(normal freezing point for acetic acid = 16.60 dg C, boiling point = 118.5 dg C)

f.p. = ??????

b.p. = ??????

b.p: delta T = Kb*molality = 3.07*m

f.p: delta T = Kf*molality = 3.90*m

molality = mols/kg solvent

mols = g/molar mass

Post your work if you get stuck.

## To find the freezing point and boiling point of the solution, we need to calculate the molality of the solution first.

Step 1: Calculate the molar mass of glucose (C6H12O6)

Molar mass of carbon (C) = 12.01 g/mol

Molar mass of hydrogen (H) = 1.01 g/mol

Molar mass of oxygen (O) = 16.00 g/mol

Therefore, molar mass of glucose (C6H12O6) = (6 * 12.01) + (12 * 1.01) + (6 * 16.00) = 180.18 g/mol

Step 2: Calculate the number of moles of glucose dissolved in the solution

Number of moles = mass (g) / molar mass (g/mol)

Number of moles of glucose = 5.00 g / 180.18 g/mol ≈ 0.0277 mol

Step 3: Calculate the molality of the solution

Molality (m) = moles of solute / mass of solvent (kg)

Mass of solvent = 500.0 g / 1000 = 0.500 kg

Molality = 0.0277 mol / 0.500 kg ≈ 0.0554 mol/kg

Step 4: Calculate the change in boiling point and freezing point of the acetic acid solution.

For boiling point:

ΔTb = Kb * molality

ΔTb = 3.07 * 0.0554 ≈ 0.170 dg C

Boiling point of the solution = normal boiling point of acetic acid + ΔTb

Boiling point = 118.5 dg C + 0.170 dg C = 118.67 dg C (rounded to two decimal places)

For freezing point:

ΔTf = Kf * molality

ΔTf = 3.90 * 0.0554 ≈ 0.216 dg C

Freezing point of the solution = normal freezing point of acetic acid - ΔTf

Freezing point = 16.60 dg C - 0.216 dg C = 16.38 dg C (rounded to two decimal places)

Therefore, the new freezing point and boiling point of the solution are approximately 16.38 dg C and 118.67 dg C, respectively.

## To find the new freezing point and boiling point of the solution, we need to calculate the molality of the glucose in acetic acid and use the respective colligative properties constants, Kf and Kb, of acetic acid.

First, let's calculate the number of moles of glucose:

Molar mass of glucose (C6H12O6) = 180.18 g/mol

moles of glucose = 5.00 g / 180.18 g/mol = 0.0277 mol

Next, we need to calculate the molality of the solution, which is defined as moles of solute per kilogram of solvent. In this case, the solvent is acetic acid, and it has a mass of 500.0 grams.

molality = moles of solute / kg of solvent

molality = 0.0277 mol / 0.500 kg = 0.0554 mol/kg

Now, using the given colligative property constants, Kf = 3.90 and Kb = 3.07, we can calculate the changes in freezing point and boiling point.

For freezing point depression (f.p.), the equation is:

ΔTf = Kf * molality

Substituting the values:

ΔTf = 3.90 * 0.0554 = 0.216 dg C

The new freezing point will be the normal freezing point of acetic acid (16.6 dg C) minus the depression in temperature:

f.p. = 16.6 dg C - 0.216 dg C = 16.384 dg C

For boiling point elevation (b.p.), the equation is:

ΔTb = Kb * molality

Substituting the values:

ΔTb = 3.07 * 0.0554 = 0.170 dg C

The new boiling point will be the normal boiling point of acetic acid (118.5 dg C) plus the elevation in temperature:

b.p. = 118.5 dg C + 0.170 dg C = 118.67 dg C

Therefore, the new freezing point is approximately 16.384 dg C and the new boiling point is approximately 118.67 dg C.