Crude oil pumped out of the ground may be accompanied by formation water, a solution that contains high concentrations of nacl and other salts. If the boiling point of a sample of formation water is 2.30°c above the boiling point of pure water, what is the molality of particles in the sample?

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Respuesta :

nuuk nuuk · Answer 1 · 2018-09-25T07:29:57+02:00

solution:

the change in the boiling point is given as,

dTbp =2.30°c

elevation constant for the solvent is given by,

kb=0.512°c/m

[tex]molality=\frac{dTBP}{KB\times m}\\

=\frac{2.30}{0.512c/m}[/tex]

= 4.49m

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nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-03-15T17:13:49+01:00

4.49m is the molality of NaCl and other salts particle in the sample.

Molality is defined as no. of moles of solute present in per kilogram of solution.

For this question, molality can be calculated as:

m = ΔTb / Kb, where

ΔTb = elevation in boiling point = 2.30°C (given)

Kb = molal elevation constant of pure water = 0.512°C/mole

Now we put these values in the above equation, we get

m = 2.30°C / 0.512°C/m = 4.49 m

Hence, 4.49m is the molality of particles.

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https://brainly.com/question/1370684