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?

Question

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?

Answer 1

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

Answer 2

To answer these questions, we first need to calculate the amount of sodium and liquid mercury used in the solution.

For Sodium (Na):

Given: Density of Na(s) = 0.97 g/cm³
Volume of Na(s) used = 1.0 cm³

Therefore, the mass of Na used can be calculated by multiplying the volume by the density:
Mass of Na = Volume of Na × Density of Na(s)
Mass of Na = 1.0 cm³ × 0.97 g/cm³
Mass of Na = 0.97 g

For Liquid Mercury (Hg):

Given: Density of Hg(l) = 13.6 g/cm³
Volume of Hg(l) used = 20.0 cm³

The mass of Hg can be calculated similarly:
Mass of Hg = Volume of Hg × Density of Hg(l)
Mass of Hg = 20.0 cm³ × 13.6 g/cm³
Mass of Hg = 272 g

Now, let's proceed to answer the questions:

(a) To calculate the molality of Na in the solution:
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

First, convert the mass of Na(s) to moles using its molar mass:
Molar mass of Na = 22.99 g/mol
Moles of Na = Mass of Na / Molar mass of Na
Moles of Na = 0.97 g / 22.99 g/mol
Moles of Na = 0.0422 mol

As we need to consider the solvent (liquid mercury) as the reference, we'll convert the mass of Hg to kilograms:
Mass of Hg = 272 g = 0.272 kg

Molality of Na = Moles of Na / Mass of Hg (in kg)
Molality of Na = 0.0422 mol / 0.272 kg
Molality of Na ≈ 0.155 mol/kg

(b) To calculate the molarity of Na in the solution:
Molarity (M) is defined as the number of moles of solute per liter of solution.

The total volume of the solution is given as 21.0 cm³. Convert this to liters:
Volume of solution = 21.0 cm³ = 0.021 L

Molarity of Na = Moles of Na / Volume of solution (in L)
Molarity of Na = 0.0422 mol / 0.021 L
Molarity of Na ≈ 2.01 mol/L

(c) Now, let's compare the molality and molarity values:
Molality of Na ≈ 0.155 mol/kg
Molarity of Na ≈ 2.01 mol/L

Since the density of the solvent (liquid mercury) is significantly higher than that of water, the molality and molarity values are not nearly equal in this case. For dilute aqueous solutions, where water is the solvent, the molality and molarity values are generally close in value due to the similar densities of the solute and solvent. However, in this sodium amalgam solution, the densities of sodium and liquid mercury differ greatly, resulting in a significant difference between the two concentrations.