Under what pressure is chloroform at a mercury reading of #940*"Torr"#?

1 Answer
Jun 16, 2016

Answer:

This is a very poor question, and has been asked by someone who has never used a mercury manometer.

Explanation:

It is a fact that #1# #atm# pressure will support a column of mercury #760# #mm# high. Have you seen a mercury manometer? Mercury has all but disappeared from modern laboratories, on the basis that it is a 100% pain in the posterior to clean up the mercury spills that inevitably occur in the laboratory.

For low pressures, a measurement in #mm# of mercury is appropriate. For pressures higher than 1 atmosphere it is completely inappropriate.

If you convert the mercury measurement to atmospheres, i.e #"940 Torr"# #=# #940/760*atm#, you have your pressure measurement. Now you have to make another assumption: that all the chloroform is present as a gas. Chloroform has a normal boiling point of #61# #""^@C#. It is unreasonable to propose that it could be present as a gas (by the way, when I say normal boiling point, I mean that its vapour pressure is equal to #1# #atm# at this temperature).

So, as you probably have already realized, the question is a real dog's breakfast. Students can offer bad answers, and examiners can equally propose bad questions, and whoever set this deserves a kick up the backside. I presume this is a first year undergraduate/2nd year phys chem problem?