For Your Initial Post, Pick A Combination Of One Of The ✓ Solved

For your initial post pick a combination of one of the

For your initial post pick a combination of one of the following molecules, one mass quantity, and one total solution volume from this table. You may not use the same combination as another student so be sure to read any existing peers postings before posting your own.

  • NaCl: 2 g, 150 ml
  • CaBr2: 6 g, 3 L
  • Li3N: 1 g, 500 ml
  • CaO: 7 g, 14 L

Then, give each of the following for your unique combination being sure to show all of your work:

  • %mass/volume concentration
  • Molarity
  • Osmolarity

Make sure to post at least two high quality posts.

Paper For Above Instructions

In this assignment, I have chosen a unique combination of molecules and solution properties from the given table. The selected combination is 2 grams of NaCl (sodium chloride) in 150 ml of solution. The aim of this post is to calculate the % mass/volume concentration, molarity, and osmolarity of this solution.

% Mass/Volume Concentration

The % mass/volume concentration is calculated using the formula:

% mass/volume = (mass of solute (g) / volume of solution (ml)) × 100

For the selected combination, the mass of NaCl is 2 g and the volume of solution is 150 ml. Plugging in these values gives:

% mass/volume = (2 g / 150 ml) × 100 = 1.33%

Thus, the % mass/volume concentration of the NaCl solution is 1.33%.

Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. To calculate molarity, we first need to find the number of moles of NaCl in 2 grams.

The molar mass of NaCl can be calculated as follows:

  • Na (Sodium): 22.99 g/mol
  • Cl (Chlorine): 35.45 g/mol

The total molar mass of NaCl is:

22.99 g/mol + 35.45 g/mol = 58.44 g/mol

To find the number of moles of NaCl in 2 grams, we can use the formula:

Number of moles = mass (g) / molar mass (g/mol)

Using this formula, we calculate:

Number of moles of NaCl = 2 g / 58.44 g/mol ≈ 0.0342 moles

Next, we convert the volume from ml to liters:

150 ml = 0.150 L

Now we can calculate the molarity:

Molarity (M) = moles of solute / volume of solution (L)

Molarity = 0.0342 moles / 0.150 L ≈ 0.228 M

Thus, the molarity of the NaCl solution is approximately 0.228 M.

Osmolarity

Osmolarity is a measure of the total concentration of solute particles in a solution. For ionic compounds like NaCl, we must consider dissociation in water. Sodium chloride dissociates into two particles: Na+ and Cl-.

Thus, for each mole of NaCl, we produce two osmoles:

Osmolarity = Molarity × number of particles

Using the molarity we calculated earlier:

Osmolarity = 0.228 M × 2 = 0.456 Osm

The osmolarity of the NaCl solution is therefore approximately 0.456 Osm.

Conclusion

In summary, for the combination of 2 grams of NaCl in 150 ml of solution, I calculated the following:

  • % mass/volume concentration: 1.33%
  • Molarity: 0.228 M
  • Osmolarity: 0.456 Osm

These calculations demonstrate the method for determining key concentration metrics for solute solutions. Understanding these concepts is vital in various applications in chemistry and biochemistry.

References

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  • American Chemical Society. (2018). Molecular Weight of Salts. Retrieved from https://www.acs.org
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