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

Jun 16, 2016

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 $1$ $a t m$ pressure will support a column of mercury $760$ $m m$ 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 $m m$ of mercury is appropriate. For pressures higher than 1 atmosphere it is completely inappropriate.

If you convert the mercury measurement to atmospheres, i.e $\text{940 Torr}$ $=$ $\frac{940}{760} \cdot a t m$, 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 $1$ $a t m$ 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?