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kifflom [539]
3 years ago
6

What is the boiling point of a 0.75-molal solution of a non-volatile non-electrolyte solute in water? Kb for water = 0.52 degree

s C/ molal.
99.61 °C
100.0 °C
100.39 °C
101.44 °C
Chemistry
2 answers:
Tom [10]3 years ago
4 0
The answer is C. I just did this assignment
AVprozaik [17]3 years ago
3 0
Take the molality and multiply it by the Kb
0.75 × 0.52 = 0.39 C

then add the boiling pt to the 0.39
100 + 0.39 = 100.39 C
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The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

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