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Topic: Boiling point elevation question  (Read 1821 times)

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Offline Danelle Vivier

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Boiling point elevation question
« on: June 08, 2013, 12:04:14 PM »
Hi, this is my first post.

The following Q was in our Engineering Chemistry exam - Stellenbosch University, South Africa

I have tried to do some calculations to find values I can use straight into the ΔTb = Kbm  formula, but I could not figure out what to do and what is the correct steps. Please help.

Question: 10 marks

The vapor pressure of an aqueous glucose (C6H12O6) solution in 17.07 mmHg at 20°C, while that of pure water is 17.25 mmHg at the same temperature. Glucose can be assumed to be non-volatile. Calculate the boiling point of the solution. The boiling point of pure water is 100°C, and the boiling point elevation constant for water is 0.512°C/m.

(m is molality)

I hope someone will be able to help. Please show steps.
Thank you very much.

Offline Borek

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Re: Boiling point elevation question
« Reply #1 on: June 08, 2013, 12:18:34 PM »
Use the vapor pressure change to calculate concentration of the glucose.
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Offline Danelle Vivier

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Re: Boiling point elevation question
« Reply #2 on: June 08, 2013, 02:45:48 PM »
Ok thanks.

I used the change in vapor pressure and got the mole fractions of glucose and water..
I took 1L of water, calculated the moles and with that calculated the moles of glucose.

Calculated molality with the moles of glucose divided by 1L of water, and got 7.7m.

Using the formula in first post, I got 3.94°C, thus the new boiling point is 103.94°C.

Is this correct?

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