What mass of glycerin (C3H8O3), a nonelectrolyte, must be dissolved in 400.0 g water in order to give a solution with a boiling point of 103.50 C?

delta T = Kb*m

Solve for molality = m

m = mols/kg solvent
Solve for mols.

mols = grams/molar mass. You know mols and molar mass, solve for grams.

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To determine the mass of glycerin (C3H8O3) needed to form a solution with a boiling point of 103.50°C, we can use the concept of boiling point elevation.

The boiling point elevation (ΔTb) is given by the equation:

ΔTb = Kb * m

Where:
ΔTb is the change in boiling point
Kb is the molal boiling point elevation constant
m is the molality of the solution (moles of solute per kilogram of solvent)

In this case, since glycerin (C3H8O3) is a nonelectrolyte and does not dissociate into ions in water, we can assume that the molality (m) is the same as the molarity (M).

First, we need to calculate the molality of the solution. We know that the solution contains 400.0 g of water, so the mass of the solvent is 400.0 g. To calculate the molality, we need to convert the mass of water to moles:

m(solvent) = mass(solvent) / molar mass(solvent)

The molar mass of water (H2O) is 18.02 g/mol, so:

m(solvent) = 400.0 g / 18.02 g/mol = 22.20 mol

Next, we need to calculate the change in boiling point (ΔTb). Given that the boiling point elevation is 103.50°C, the change in boiling point is:

ΔTb = Tb(solution) - Tb(solvent)

where Tb(solution) is the boiling point of the solution and Tb(solvent) is the boiling point of the pure solvent (water). In this case, Tb(solvent) is 100.0°C, as that is the boiling point of water at standard conditions.

ΔTb = 103.50°C - 100.0°C = 3.50°C

Finally, we can use the equation ΔTb = Kb * m to solve for the molality (m) of the solution:

3.50°C = Kb * 22.20 mol

Rearranging the equation:

Kb = 3.50°C / 22.20 mol

At this point, we need the molal boiling point elevation constant (Kb) for water, which we can find in a reference table. The molal boiling point elevation constant for water is approximately 0.512°C/m. Substituting this value into the equation:

0.512°C/m = 3.50°C / 22.20 mol

Now, we can solve for the molality (m):

m = 3.5°C / (22.20 mol * 0.512°C/m)

m ≈ 0.0651 mol/kg

Finally, we need to calculate the mass of glycerin (C3H8O3) required to give a molality of 0.0651 mol/kg in 400.0 g of water.

mass(solute) = m * molar mass(solute) * mass(solvent)

The molar mass of glycerin (C3H8O3) is:
molar mass(C3H8O3) = (3 x molar mass(C)) + (8 x molar mass(H)) + (3 x molar mass(O))
= (3 x 12.01 g/mol) + (8 x 1.008 g/mol) + (3 x 16.00 g/mol)
= 92.09 g/mol

Substituting the values into the equation:

mass(solute) = 0.0651 mol/kg * 92.09 g/mol * 400.0 g

mass(solute) ≈ 2.42 g

Therefore, approximately 2.42 grams of glycerin (C3H8O3) must be dissolved in 400.0 g of water to give a solution with a boiling point of 103.50°C.

To find the mass of glycerin that must be dissolved in water to give a solution with a boiling point of 103.50 °C, we need to use the formula for boiling point elevation.

The formula for boiling point elevation is given by:

ΔTb = Kb * m

where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point elevation constant for the solvent (water), and
m is the molality of the solute in the solution.

First, we need to calculate the boiling point elevation (ΔTb). The boiling point of pure water is 100 °C, so the elevation in temperature is ΔTb = 103.50 °C - 100 °C = 3.50 °C.

Next, we need to find the molal boiling point elevation constant (Kb) for water. The literature value for Kb of water is 0.512 °C/m.

Now we can rearrange the formula to solve for the molality (m):

m = ΔTb / Kb

Substituting the values:

m = 3.50 °C / (0.512 °C/m) = 6.84 m

The molality (m) is 6.84 mol/kg.

To find the mass of glycerin, we need to convert the molality (m) to moles of glycerin:

moles of glycerin = molality (m) * mass of water

The mass of water is given as 400.0 g.

So, moles of glycerin = 6.84 mol/kg * 400.0 g = 2736 g

Therefore, the mass of glycerin that must be dissolved in 400.0 g of water is 2736 grams.