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?

determine the final concentration of a 20.ml 5.M NaCl solution if 30Mml of water has is added to the solution?

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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.

To calculate the molality of Na in the solution, we need to find the moles of Na and the mass of Hg.

(a) To find the moles of Na, we can use its density and volume:
Moles of Na = Density of Na * Volume of Na
= 0.97 g/cm3 * 1.0 cm3
= 0.97 g

(b) To find the mass of Hg, we can use its density and volume:
Mass of Hg = Density of Hg * Volume of Hg
= 13.6 g/cm3 * 20.0 cm3
= 272 g

Now, we can calculate the molality of Na:
Molality = Moles of Na / Mass of Hg
= 0.97 g / 272 g
= 0.00357 mol/g

To calculate the molarity of Na in the solution, we need to convert the volume of Na from cm3 to liters, and then divide the moles of Na by the volume in liters.

(c) Molarity = Moles of Na / Volume of Na in liters
= 0.97 g / 21.0 cm3 * (1 L / 1000 cm3)
= 0.000046 mol / 0.021 L
= 0.00219 mol/L

Comparing the molality and molarity values, we can see that they are not nearly equal in value. The molality is 0.00357 mol/g, while the molarity is 0.00219 mol/L. Therefore, for the sodium amalgam described here, the molality and molarity are not nearly equal.