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?
1 answer
I think I answered this question for you earlier. (A day or so ago). Look at your earlier posts.
0 answers