Sodium metal dissolves in liquid mercury to form a solution called a sodium amalgam. The densities of Na(s) and Hg(l) are 0.97 g/cm3 and 13.6 g/cm3, respectively. A sodium amalgam is made by dissolving 1.0 cm3 Na(s) in 20.0 cm3 Hg(l). Assume that the final volume of the solution is 21.0cm3. (a) calculate the molality of Na in the solution. (b) calculate the molarity of Na in the solution. (c) for dilute aqueous solutions, the molality and molarity are generally nearly equal in value. Is that the case for sodium amalgam described here?
Chemistry - pk, Sunday, December 11, 2011 at 9:15am
determine the final concentration of a 20.ml 5.M NaCl solution if 30Mml of water has is added to the solution?
Chemistry---to pk - DrBob222, Sunday, December 11, 2011 at 2:35pm
Please don't muddle up the board with duplicate questions ESPECIALLY piggy backing on another. Go to the top of the page and post your own question. You'll get faster service that way.
Chemistry--to bernan - DrBob222, Sunday, December 11, 2011 at 2:42pm
Find grams Na and grams Hg.
g Na = 0.97 g/cc x 1 cc = 0.97g
g Hg = 13.6 g/cc x 20 cc = about 270 g (all of this is about. You should redo the math on your calculator. It will be better than my estimates.)
molality = moles Na/kg solvent.
moles Na = g Na/molar mass Na.
kg solvent = 0.270
Substitute and solve for m.
For molarity. M = moles/L soln
moles Na you calculated above.
L soln = 21 cc = 0.021 L.
Substitute into M = moles/L and calculate.
The c part you can answer after you've completed a and b parts.
Post your work if you get stuck.