posted by Amy .
At room temperature pure water has a density of 1.00 g/ mL. What mass of water (in grams) will fit into a container if 1.00 lb. of mercury completely fills the container?
Okay, here is what I solved so far.
Let's say H2o equals water and Hg equals mercury= 13.6g/mL. equals an expression for the density of mercury and h2o is expressed 1.00 g\ mL.
1. 13.6/mL. * 1lb./454 grams (equals grams in 1lb.) * 1,000 mL\ Liters * Liter\ 3.785 g.
Please help me! I am a beginning student in Chemistry. Until then, I will carefully review what is being taught in text and try to patiently solve the problems (I have two other questions related to this subject).
Here is what I would do. I think it will make it a little simpler than trying to convert density units.
1 lb Hg x (454 g/lb) = 454 grams Hg.
volume of container which holes the Hg is
volume = mass/density = 454/13.6 = ?? mL.
Then go to the water medium.
massH2O = volumeH2O x density H2O
massH2O = xx mL x densityH2O.
Solve for mass H2O.
The thing you said about the volume of the container didn't make sense. So far, I only have the the density for H2O and Hg. So I couldn't put that the volume of the container was 454/13.6. Density over density?
The volume of the container is unknown but the density and mass of Hg is given so you can calculate the volume. Note that the volume when Hg is in the container and when water is in the container is the same. The volume is the volume is the volume. (Side note: density/density doesn't make sense to me either but 454/13.6 is not density--it's the volume which the Hg occupies and it tells you the volume of the container.)
mass of the Hg container = volume Hg x density Hg.
Solve for volume of Hg.
vol Hg = mass Hg/density Hg
vol Hg = 454 g Hg/13.6 = 33.38 mL; therefore, the Hg container holds 33.38 mL.
How much water (what mass) can be placed in that container (that has a volume of 33.38 mL)?
massH2O = volume H2O x density H2O.
mass H2O = 33.38 mL x 1.00 g/mL = 33.38 mL H2O, then I would round to 3 places to give a final answer of 33.4 grams.
You've been very helpful (if only I could thank you by name). For now, I will just say, "Thanks Bob."
Now, 2/3 of my problems are understood. I may be able to solve the last one alone.