Are my calculations correct for this question?

[bamdavis]

Must find the molar mass of an unknown volatile liquid.

 

Using the equation PV=mRT

M

 

P= 102.4 KPA

V=0.098L

T= 373 K

Mass= 0.1647 g

R= constant gas for KPA= 8.312 KPA

 

Rearrange it to M= mRT

PV

 

i got M= (0.1647)(8.312)(373)

(0.098)(102.4)

 

M= 50.8841 g/mol

[Riogho]

I don't know what you mean by volatile liquid, but in my experience using the Ideal Gas Law for a liquid is not going to give you a very accurate answer.

[Cap'n Refsmmat]

Indeed. Mind explaining your reasoning, bamdavis?

[bamdavis]

this is just from a lab i did today.

we put 5ml of an unknown volatile in a volumetric flask, put tin foil on the top and poked a hole in it. We heated it in a beaker filled with boiling water to get it to evaporate. After it evaporated we cooled it to condense into a liquid and then weighed it.

[hermanntrude]

OK so the ideal gas equation is fine to use while the liquid was vapourised.

[bamdavis]

I dont understand

[Cap'n Refsmmat]

The ideal gas law only works on gases. That means you can't use it on a bottle of water or a flask full of water. But you can use it on steam -- vaporised water. Steam is a gas, so the ideal gas law applies.

[Riogho]

Let's put it this way buddy, when you worked out your equation, the element you got was Vanadium. Which is seriously doubt is a liquid at 373K.

[hermanntrude]

it doesn't have to be an element, riogho.

 

I havent checked bamdavis's math is correct, but, assuming he is, a mass of 50.9 g/mol is perfectly feasible. it's possible that it's chloromethane (which has a mass of 50.5 ish), for instance... although that's a fairly uncommon solvent for a school lab... i suspect it's not that, but there are many others it could be, especially if you account for the fairly wide range of possible experimental error

[macscigirl]

I thought R had to be .08206

[Cap'n Refsmmat]

Depends on what units you're using.

[hermanntrude]

I thought R had to be .08206

 

R is 0.08206 [math]L.atm.mol^{-1} K^{-1}[/math]

 

it's ALSO 8.3145 [math]L.kPa.mol^{-1}K^{-1}[/math]

 

if you put the units into the calculation above you'll see that the second version is the appropriate one to use.

 

There also other values for R. For instance you could have the units [math]L.mmHg.mol^{-1}K^{-1}[/math]

 

or you could have the units [math]cm^{3}atm.mol^{-1} K^{-1}[/math]

 

etc etc etc