Concentration Units and Solutions – Comprehensive Study Notes

Key Definitions

• Solution – homogeneous mixture of two or more substances (solute + solvent).
• Concentration – quantitative expression of “how much solute is present” in a given amount of solution or solvent.
• Solute – substance being dissolved; Solvent – substance doing the dissolving (larger amount).

Molarity (symbol: M, units: $\text{mol\,L}^{-1}$)

• Formal definition: $M = \dfrac{n<em>{\text{solute}}}{V</em>{\text{solution}}}$
– $n<em>{\text{solute}}$ = moles of solute. – $V</em>{\text{solution}}$ = total volume of solution in litres (L).
• Re-arrangements frequently used:
– Moles from volume & molarity: $n = M \times V$
– Volume from moles & molarity: $V = \dfrac{n}{M}$
• Units may be written “$\text{mol}\,/\,\text{L}$” or capital “M.” Example: “0.25 M.”

Calculations Involving Molarity

1 – Determining molarity from a given mass of solute

Example (from video):
• Data: 5.85 g NaCl dissolved to make 250 mL solution.
• Steps

1. Convert grams → moles using molar mass $M_r(\text{NaCl}) \approx 58.44\,\text{g mol}^{-1}$: $5.85\,\text{g}\times\dfrac{1\,\text{mol}}{58.44\,\text{g}} = 0.100\,\text{mol}$
2. Convert volume to litres: $250\,\text{mL} = 0.250\,\text{L}$.
3. Apply definition: $M = \dfrac{0.100\,\text{mol}}{0.250\,\text{L}} = 0.400\,\text{M}$.

2 – Determining required mass for a target molarity & volume

Example: “What mass of KBr is needed to prepare 100 mL of 0.25 M solution?”
• Convert volume: $100\,\text{mL}=0.100\,\text{L}$.
• Find moles: $n = M \times V = 0.25\,\text{mol L}^{-1}\times0.100\,\text{L}=0.0250\,\text{mol}$.
• Mass: $m = n\times M_r(\text{KBr})\;(119.0\,\text{g mol}^{-1}) = 0.0250\,\text{mol}\times119.0\,\text{g mol}^{-1} = 2.98\,\text{g}$.

3 – Finding volume when given mass & molarity

Example: “Volume of 1.25 M solution containing 3.00 g LiOH.”
• Molar mass $M_r(\text{LiOH}) \approx 23.95\,\text{g mol}^{-1}$.
• Moles: 3.00\,\text{g}\times\dfrac{1\,\text{mol}}{23.95\,\text{g}} = 0.125\,\text{mol (≈)}}.
• Volume: $V = \dfrac{n}{M} = \dfrac{0.125\,\text{mol}}{1.25\,\text{mol L}^{-1}} = 0.100\,\text{L}$ ≈ 100 mL.

Dilutions

• Dilution – adding solvent to a solution → concentration decreases; total moles of solute remain constant.
• Equality of moles before (1) and after (2) dilution: $n<em>1 = n</em>2$.
– Since $n = C \times V$, we obtain the classic dilution equation: $C<em>1V</em>1 = C<em>2V</em>2$ (often written $M<em>1V</em>1=M<em>2V</em>2$).
– Works with any concentration/volume units, provided the units match on both sides.
– NOT valid for chemical reactions (no stoichiometric ratios included).

Worked examples

1. Volume of 2.0 M stock to make 4.0 L of 0.50 M solution:
   $V<em>1 = \dfrac{C</em>2V<em>2}{C</em>1} = \dfrac{0.50\,\text{M}\times4.0\,\text{L}}{2.0\,\text{M}} = 1.0\,\text{L}$.
   – Practical view: measure 1.0 L of stock, transfer to volumetric flask, add solvent up to 4.0 L mark.
2. Concentration of stock if 2.00 mL diluted to 10.0 mL gives 0.25 M:
   $C<em>1 = \dfrac{C</em>2V<em>2}{V</em>1} = \dfrac{0.25\,\text{M}\times10.0\,\text{mL}}{2.00\,\text{mL}} = 1.25\,\text{M}$.

Alternative Concentration Units

Percent by Mass (% w/w)

• Definition: $\%\,w/w = \dfrac{\text{mass solute}}{\text{mass solution}}\times100$.

Percent by Volume (% v/v)

• Definition: $\%\,v/v = \dfrac{\text{volume solute}}{\text{volume solution}}\times100$.

Parts-per-million (ppm) & Parts-per-billion (ppb)

• ppm: multiply the fraction by $1.0\times10^6$.
• ppb: multiply the fraction by $1.0\times10^9$.
• Used for very dilute solutions (environmental chemistry, trace contaminants).

Molality (symbol: m, units: $\text{mol}\,\text{kg}^{-1}$)

• Definition: $m = \dfrac{n<em>{\text{solute}}}{m</em>{\text{solvent}}(\text{kg})}$.
• Key distinctions from molarity:
– Denominator is mass of solvent (kg), not volume of solution.
– Unaffected by temperature or pressure because masses do not change with thermal expansion.
– Preferred for colligative property calculations (boiling-point elevation, freezing-point depression, osmotic pressure).
• Notation warning: lowercase “m” (molality) versus uppercase “M” (molarity).

Practical & Conceptual Connections

• Laboratory Reality: You rarely measure moles directly; you weigh solids or measure volumes, then convert.
• Unit Conversions: Always ensure volumes are in litres for molarity; kilograms for molality.
• Stock Solutions: Buying or preparing a high-concentration stock and diluting saves time and storage space.
• Equality of Moles Principle: Underpins both rearranged molarity equations and the dilution equation.
• Difference in Focus: Molarity/percent/ppm emphasise composition of whole solution; molality isolates solvent properties.
• Ethical & Safety Implications: Accurate concentration calculations ensure correct dosages (pharmaceuticals), comply with environmental limits (heavy-metal ppm levels), and prevent hazardous over-concentration in lab reactions.

Common Pitfalls & Tips

• “mL vs L” errors are the most frequent source of 1000× mistakes—always confirm units before final answer.
• Double-check molar masses to appropriate significant figures.
• Remember that dilution equation is invalid for stoichiometric reaction problems.
• Distinguish lowercase m (molality) and uppercase M (molarity) both in writing and when reading literature.
• When switching temperature, molarity can change (due to solution expansion); molality will not—choose accordingly.

Quick Reference Formula List

• Molarity: $M = \dfrac{n}{V<em>{(L)}}$. • Moles: $n = M\,V</em>{(L)}$.
• Mass from molarity: $m = M<em>r\,M\,V</em>{(L)}$.
• Dilution: $C<em>1V</em>1 = C<em>2V</em>2$.
• Molality: $m = \dfrac{n}{m<em>{\text{solvent}}(kg)}$. • Percent by mass: $\%w/w = \dfrac{m</em>{solute}}{m<em>{solution}}\times100$. • Percent by volume: $\%v/v = \dfrac{V</em>{solute}}{V_{solution}}\times100$.
• ppm / ppb: $\text{ppm} = \dfrac{\text{mass solute}}{\text{mass solution}}\times10^6$; $\text{ppb} = \times10^9$.

With these principles, equations, and worked examples, you are prepared to tackle preparation, dilution, and analysis of almost any laboratory solution. Remember—careful unit handling and clear problem-solving steps are the keys to success.