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lesya692 [45]
3 years ago
5

Calculate the new boiling point of a solution if 10.00 g of a non-ionizing compound (C3H5(OH)3) is dissolved in 90.00 g of H2O.

Molar Mass of C3H5(OH)3 = 92.09 g/mol Kb = 0.51 oC/m (Answer must be in 4 sig fig. Do not include units in your answer).
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
5 0

Answer:

Boiling T° of solution = 100.6

Explanation:

Formula for elevation of boiling point is:

ΔT = Kb . m . i

where ΔT means Boiling T° of solution - Boiling T° of pure solvent

Our solute is a non ionizing compound.

i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.

m = molality (moles of solute dissolved in 1 kg of solvent)

90 g of solvent = 0.09 kg of solvent

We convert mass of solute to moles (by the molar mass):

10 g . 1 mol /92.09 g = 0.108 moles

m = 0.108 mol /0.09 kg = 1.21 m

Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1

Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C

Boiling T° of solution = 100.6

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What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
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