Boiling Point Chem

Question

0.7 mol of table sugar(C^12H622O^11) is dissolved in 350g of water(Kb=0.51 P H^2O=100.00

How would I find the boiling point?

Answer 1

To find the boiling point elevation, we can use the following equation:

ΔTb = Kb × m

where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant (0.51 °C/m for water), and m is the molality of the solution (moles of solute per kilogram of solvent).

First, let's calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

We are given that there are 0.7 mol of table sugar (C12H22O11) and 350 g of water. To convert the mass of water to kilograms, we divide by 1000:

mass of water = 350 g / 1000 = 0.35 kg

molality = 0.7 mol / 0.35 kg = 2.0 mol/kg

Now we can use the boiling point elevation equation:

ΔTb = Kb × m

ΔTb = 0.51 °C/m × 2.0 mol/kg = 1.02 °C

Therefore, the boiling point elevation of the solution is 1.02 °C.

To find the boiling point of the solution, we can use the following equation:

Tb = Tbo + ΔTb

where Tb is the boiling point of the solution, Tbo is the boiling point of the pure solvent (which is 100.00 °C for water), and ΔTb is the boiling point elevation we just calculated.

Tb = Tbo + ΔTb

Tb = 100.00 °C + 1.02 °C = 101.02 °C

Therefore, the boiling point of the solution is 101.02 °C.