I really need to help with this Chemistry problem!?

Question

Attempt to determine the density of the unknown from the following data:

Mass of dry pycnometer & stopper : 32.4245g
Mass of pycnometer + stopper + H2O: 58.0558g
Temp. of H2O: 23.0°C
Mass of pycnometer + stopper + unknown liquid: 52.8734g

(a) Find density of H2O @ 23.0°C.
(b) Calculate the volume of the pycnometer.
(c) Determine the density of your unknown liquid.

Answer 1

Well, I saw this question a hour ago, and I supposed somebody else would answer, I didn’t have time; I see that you still need help, so here it is:

You determined steps well:

First we will find density of water at 23 degrees; I found at http://www.simetric.co.uk/si_water.htm it is 0.997538 g/cm3

Then, we are going to calculate volume of picnometer, using its data for water; mass of water is, obviously, difference in masses of dry picnometer with stopper and mass of picnometer with stopper and with water; so it is:

mass of water = 58.0558 – 32.4245 = 25.6313 grams

From here we calculate volume of picnometer as:

volume = mass of water / density of water = 25.6313 / 0.997538 = 25.69456 cm3

Now we need a mass of unknown liquid, similarly to mass of water:

mass of liquid = 52.8734 – 32.4245 = 20.4489 grams

And finally, its density is:

density = mass / volume = 20.4489 / 25.69456 = 0.7958 g/cm3

Hope this was understandable, and not too late… It is a standard way of calculation of density using the picnometer. You understand, the volumes of water and unknown liquid are the same- the volume of picnometer.