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Topic: Chemistry freezing point and determination of molecular formula  (Read 4839 times)

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Offline Forceofmatter

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question 1: Calculate the mole fraction of benzene in benzene-toluene solution that is in equilibrium at 25 deg. Celsius with a vapor pressure that contains 62% mol Benzene. At 25 deg. Celsisus the vapor pressures of benzene and toluene are 95.1 mm Hg and 28.4 mm Hg respectively. Kf of Benzene= 5.50 deg Celsius.

question 2: Coniferin is a glycoside (derivative of sugar) found in conifers such as fir trees. When a 1.205 g smaple of coniferin is subjected to combustion analysis, the products are 0.698 g H20 and 2.479 g CO2. In another experiment, a 2.216 g sample of coniferin is dissolved in 46.68 g water, and the normal boiling point of the solution is 100.068 deg. Celsius. What is the molecular formula of coniferin?

Help on one or both of these questions would be greatly appreciated. Please provide explanation or mathematical work showing how you got to your answer.

Offline Borek

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Re: Chemistry freezing point and determination of molecular formula
« Reply #1 on: March 30, 2009, 04:25:14 PM »
Help on one or both of these questions would be greatly appreciated. Please provide explanation or mathematical work showing how you got to your answer.

Please read forum rules.
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Offline Forceofmatter

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Re: Chemistry freezing point and determination of molecular formula
« Reply #2 on: April 02, 2009, 08:17:58 PM »
question 1: Calculate the mole fraction of benzene in benzene-toluene solution that is in equilibrium at 25 deg. Celsius with a vapor pressure that contains 62% mol Benzene. At 25 deg. Celsisus the vapor pressures of benzene and toluene are 95.1 mm Hg and 28.4 mm Hg respectively. Kf of Benzene= 5.50 deg Celsius.

I have no idea what to do with the 62%. If I knew how to solve total pressure of solution, it would be just plugging into raoult's law:
Psoln=P molfract (A) + P molfract (B).
question 2: Coniferin is a glycoside (derivative of sugar) found in conifers such as fir trees. When a 1.205 g smaple of coniferin is subjected to combustion analysis, the products are 0.698 g H20 and 2.479 g CO2. In another experiment, a 2.216 g sample of coniferin is dissolved in 46.68 g water, and the normal boiling point of the solution is 100.068 deg. Celsius. What is the molecular formula of coniferin?

I solved for an empirical formual by dividing both CO2 and H20 by their molar masses, but after doing that I could not get a conclusive answer because little changes in method would get me different empirical formulas. I know what to do with the 2nd part, find molality and use this to find molar mass of molecular formula. 
« Last Edit: April 02, 2009, 09:02:26 PM by macman104 »

Offline Borek

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Re: Chemistry freezing point and determination of molecular formula
« Reply #3 on: April 03, 2009, 04:03:50 AM »
Do you know how to calculate partial pressure of benzene over the mixture for a given benzene molar fraction?

What is the molar mass of coniferine that you have found?
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