Cliffs AP Chemistry, 3rd Edition

The students opened the flask and thus made the pressure inside the flask equal to the pressure
outside it.

(b) Restatement: Mass of vapor inside the flask.

63.088 g −62.371 g = 0.717 g

(c) Restatement: Mass of 1 mole of vapor.

PV nRT PV

gRT

" MW

::

==

PV

gRT

MW :

::

=

..

...

0 964 0 2619

0 717 0 0821 374 1

mole K atm liter

g liter atm K

::

:::

=

__

_ __

ii

i ii

= 87.2 g/mole

(d) Restatement: How could each of these mistakes affect the molecular mass of the liquid?

(1) The volatile liquid contained nonvolatile impurities.

The molecular mass would be too highbecause the nonvolatile impurities would

contribute additional mass. The contribution to volume would be negligible.

(2) The students removed the flask from the water bath before all the liquid had

vaporized.

The mass of the condensed vapor would be too high. The excess mass would be due

to both the vapor and the mass of liquid that had not vaporized. Examine the equation

for determining the molecular weight.

PV

gRT

MW :

::

=

The value for g would be too high, so the calculated MW would also be too high.

(3) There was a hole in the stopper.

The mass of the vapor would be too small, because some of the vapor would have es-

caped through the hole in the stopper before condensing on the flask. The variable g

would be smaller than expected, resulting in a molecular mass lower than expected.

(4) A few drops of water were left on the flask from the water bath when the final mass

was taken.

The mass of the condensate would be too large, because it would include both the

mass of the condensate and the mass of the water left on the flask. The variable g

would be larger than expected, resulting in a molecular mass higher than expected.

Part II: Specific Topics