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SOLUTION IN CHEMESTRY

Part 1 — Choose and Calculate

Many medications are formulated as solutions to enhance absorption and therapeutic effectiveness. To ensure these solutions are safe and compatible with the human body, healthcare professionals must accurately calculate molarity, percent solutions, and osmolarity, especially to maintain isotonicity with blood plasma (Tonog & Lakhkar, 2022).

Select one molecule, one mass quantity, and one total solution volume from the provided table.
Solute
Molecule

Solute
Mass

Total
Solution Volume

CaBr2 1.5 g 150 mL

Li3N 2.0 g 540 mL

NaCl 6.0 g 3.5 L

CaO 7.0 g 12.0 L

Using the formulas below, calculate:
Percent solution (as percent mass/volume)
Molarity
Osmolarity
Moles = mass / molar mass
Molarity (M) = moles of solute / liters of solution
Mass/volume percent = (grams of solute / volume (in mL) of solution) × 100%
Osmolarity = molarity × number of ions in the solution
NaCl = 2 (1 Na⁺ + 1 Cl⁻)
MgCl₂ = 3 (1 Mg²⁺ + 2 Cl⁻)
Part 2 — Write and Explain

In 5-10 sentences, explain your process and results:

Describe your chosen molecule and why it’s important in healthcare or biological systems.
Show and explain your calculations clearly.
Interpret what your osmolarity value means for maintaining isotonicity.

Sample Answer

To complete this activity, I have selected Lithium nitride ($Li_3N$) with a mass of 2.0 g and a total solution volume of 540 mL.

Part 1 — Calculations

Given Values:

Solute: $Li_3N$

Mass: $2.0\text{ g}$

Volume: $540\text{ mL}$ (which is $0.540\text{ L}$)

Molar Mass of $Li_3N$: $(3 \times 6.94) + 14.01 = 34.83\text{ g/mol}$

Percent Solution (Mass/Volume %) $\text{Mass/volume percent} = \left(\frac{2.0\text{ g}}{540\text{ mL}}\right) \times 100\% = \mathbf{0.37\%}$ 2. Molarity (M)

Step A: Find Moles

$\text{Moles} = \frac{2.0\text{ g}}{34.83\text{ g/mol}} = 0.05742\text{ moles}$

Step B: Calculate Molarity

$\text{Molarity} = \frac{0.05742\text{ mol}}{0.540\text{ L}} = \mathbf{0.106\text{ M}}$

3. Osmolarity

$Li_3N$ dissociates into 4 ions ($3Li^+$ and $1N^{3-}$).

$\text{Osmolarity} = 0.106\text{ M} \times 4 = \mathbf{0.424\text{ osmol/L (or 424 mOsm/L)}}$