The boiling point of water at 756 mm Hg is 99.85 degree C. What mass percent of surcrose (C12H22O11) should be present to raise the boiling point to 100.00 degree C at this pressure?

delta T = Kb*molality.

Solve for molality.

molality = moles/kg solvent.
I would use 100 g water (0.1 kg) and solve for moles.

moles = grams/molar mass.
Solve for grams.

You now have grams in 0.1 kg solvent. You convert that to mass percent.

Why did the sugar molecule go to school? Because it wanted to raise its boiling point IQ!

But on a more serious note, let's solve this question together. To determine the mass percent of sucrose required to raise the boiling point to 100 degrees Celsius at 756 mm Hg, we need to use the equation:

ΔTb = Kb * m

Where:
ΔTb is the change in boiling point (0.15 °C in this case, as we want to raise it from 99.85 °C to 100.00 °C)
Kb is the molal boiling point elevation constant for water (0.512 °C/m)
m is the molality of the solution in moles of solute per kilogram of solvent

Since we want to find the mass percent of sucrose, we first need to convert the molality (m) into mass percent (%):

m = (moles of solute) / (mass of solvent in kg)
% mass = (mass of solute) / (mass of solution) * 100

Now, let's assume we have 1 kg of water (solvent). Since the molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol, we can calculate the moles of sucrose needed as follows:

moles of sucrose = (ΔTb * m) / Kb
mass of sucrose = moles of sucrose * molar mass

Finally, we can determine the mass percent of sucrose:

% mass = (mass of sucrose) / (mass of solution) * 100

By plugging in the values and doing the calculations, we can find the mass percent of sucrose needed to raise the boiling point to 100 degrees Celsius at 756 mm Hg.

To calculate the mass percent of sucrose (C12H22O11) needed to raise the boiling point of water to 100.00 degree C at 756 mm Hg, we can use the equation:

ΔTb = Kb · m

where:
ΔTb = elevation in boiling point (0.15 degree C in this case)
Kb = boiling point elevation constant for water (0.512 degree C/m in this case)
m = molality of the solute (sucrose) in water in mol/kg

To find the molality (m), we can use the equation:

m = moles of solute / mass of solvent (water in this case) in kg

First, let's calculate the moles of water (solvent) needed to raise the boiling point using the equation:

ΔTb = Kb · m

0.15 = 0.512 · m

m = 0.15 / 0.512
m ≈ 0.293 mol/kg

Now, we need to find the mass of sucrose (C12H22O11) needed to achieve this molality.

To calculate the mass of sucrose, we can use the equation:

mass = moles of solute × molar mass of solute

The molar mass of sucrose (C12H22O11) can be calculated as:

12C + 22H + 11O = 12.01 g/mol + 22(1.008 g/mol) + 11(16.00 g/mol)

molar mass of C12H22O11 = 342.3 g/mol

Now, we can calculate the mass of sucrose:

mass = 0.293 mol/kg × 342.3 g/mol
mass ≈ 100.2 g

Therefore, a mass percent of approximately 100.2 g of sucrose should be present to raise the boiling point of water to 100.00 degree C at 756 mm Hg.

To solve this problem, we can use the formula for boiling point elevation:

ΔT = Kbp * m

Where:
ΔT = change in boiling point
Kbp = boiling point elevation constant for the solvent
m = molality of the solute

Given that the boiling point of water at 756 mm Hg is 99.85 °C and the desired boiling point is 100.00 °C, we can calculate the change in boiling point as follows:

ΔT = 100.00 °C - 99.85 °C = 0.15 °C

The boiling point elevation constant for water is 0.52 °C/m. We need to find the molality of the solute (sucrose) that would cause this change in boiling point. Rearranging the formula, we have:

m = ΔT / Kbp

m = 0.15 °C / 0.52 °C/m
m ≈ 0.29 m

Now, we need to find the mass percent of sucrose (C12H22O11) in the solution that would result in a molality of 0.29 m.

To determine the mass percent, we need to consider the molar mass of sucrose. The molar mass of C12H22O11 is calculated as follows:

(12*12.01) + (22*1.01) + (11*16.00) = 342.34 g/mol

Now, we can calculate the mass of sucrose needed to make a 0.29 m solution:

mass = molar mass * moles
mass = 342.34 g/mol * 0.29 mol
mass ≈ 99.24 g

Therefore, to raise the boiling point of water at 756 mm Hg to 100.00 °C, a mass percent of approximately 99.24% sucrose (C12H22O11) should be present in the solution.