[melliep]

The organic compound di-n-butyl phthalate, C6H22O4 (liquid), is sometimes used as a low-density (1.046 g*mL^-1) manometer fluid. Compute the pressure (in torr) of a gas that supports a 653-mm column of di-n-butyl phthalate. The density of mercury is 13.53 g*mL^-1

We're doing PV=nRT and Avogadro's Law and P1T1 etc. etc. if that helps?

I've been doing this one problem for hours. I finished all the other questions but this one has me along with all my friends stumped. If someone could help I would be extremely grateful.

Thank you Thank you =)

EDIT: ....please?? Does anyone know how to do it? It's due in 10 minutes =S. Sorry.
 

[Borek]

This is almost direct application of the pressure definition, you don't need any of the ideal gas equations.

[MrTeo]

[...] as a low-density (1.046 g*m^-1) [...] The density of mercury is 13.53 g*m^-1

You meant to write g∙cm^-3, didn't you?

Anyway... what't the pressure of a liquid column of density $$ \delta /$$ and height $$ h /$$? You can easily find this with the definition of pressure:

$$ p=\frac{F_{\per}}{S} /$$

And what's the pressure of an Hg column 760mm high?

[melliep]

No, it's mL.

[melliep]

Nevermind, time's up.
Thanks =)

[MrTeo]

No, it's mL.

1 cm³=1 mL

Nevermind, time's up.
Thanks =)

You didn't give us much time, did you? ^^

[melliep]

Haha I know this was my last resort. Sorry it was rushed. Thank you very much =)

[billnotgatez]

is it still required to memorize 760 torr = 1 atmosphere and 1 torr = 1mm mercury