An empty vial weighs 56.33 g. The vial weighs 183.50 g when filled with liquid mercury (d= 13.53 g/cm cubed). How much would the vial weigh if it were filled with water (d= 0.997 g/cm cubed at 25 degree C)?
2007-06-10 18:43:10 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
We use the mercury data to find the volume of the vial. Then we can find the weight of the vial with water.
Call WHg = 183.50 g-56.33 g
Call Vv = WHg/13.53
Now Vial wt with Water = 56.33 + Vv * 0.997
2007-06-10 18:49:31 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
183.50 - 56.33 = mass of Hg.
use density = mass divided by volume and calculate the volume. This value will be in cm^3
use density = mass divided by volume again, but use the volume just calculated as well as the density (0.997). The answer will be the mass of water.
Add 56.33 to the mass and there is your answer.
BTW, the way the question is phrased, a case could be made for the answer to be 56.33, however we both know it means the mass of the vial plus the enclosed water.
2007-06-10 18:51:16 · answer #2 · answered by ChemTeam 7 · 0⤊ 0⤋
65.7g
183.5g - 56.33g = 127.17g (wt of Hg)
127.17g x (1cc/13.53g) = 9.399cc of Hg
the vial is made to contain 9.399cc
so if it is filled with 9.399cc of water...
9.399cc x (0.997g/cc) = 9.37g
the vial with water will weigh...
9.37 g + 56.33g = 65.7 g
2007-06-10 19:33:03 · answer #3 · answered by hey_it's_Ray 2 · 0⤊ 0⤋