Gen Chem Ch. 9 - Molarity and Dilutions

What is a solution?

A homogeneous mixture in which two or more substances are uniformly dispersed at the molecular level.

Define solvent.

The component present in the largest amount; it dissolves the solute.

Define solute.

The substance present in smaller amount that becomes evenly distributed within the solvent.

What is an aqueous solution?

A solution in which water acts as the solvent.

Why are solutions important in chemistry?

Most chemical reactions occur in solution because dissolved reactants move freely and collide effectively.

What is required for a reaction to occur?

Physical contact between reactant particles; easiest in gases and liquids.

Differentiate between dilute and concentrated solutions.

Dilute = small solute-to-solvent ratio; concentrated = large solute-to-solvent ratio.

Which contains more solute, 0.1 M or 1.0 M NaCl?

The 1.0 M solution.

Define molarity (M).

M = moles of solute / liters of solution.

What are the units of molarity?

mol L⁻¹ (read as "moles per liter").

Example: 0.100 M NaCl solution.

Contains 0.100 mol NaCl in 1.00 L solution or 0.0100 mol in 0.100 L.

How is molarity used as a conversion factor?

mol = M × V and V = mol / M.

Example: Find the molarity of 11.5 g NaOH in 1.50 L solution.

11.5 ÷ 40.00 = 0.2875 mol; 0.2875 / 1.50 = 0.192 M NaOH.

Example: How many mL of 0.250 M NaCl needed for 0.100 mol NaCl?

0.100 / 0.250 = 0.400 L = 400 mL.

Why is molarity convenient?

It directly links measurable volume to moles of solute for stoichiometric use.

Describe how to prepare a solution of known molarity.

1. Weigh solute

2. Transfer to volumetric flask

3. Add water and dissolve

4. Fill to mark 5 Invert to mix thoroughly.

Why use a volumetric flask?

It ensures the final volume is accurate, giving precise concentration.

Example: How many grams Sr(NO₃)₂ to make 250.0 mL of 0.100 M solution?

Moles = 0.100 × 0.250 = 0.0250

Mass = 0.0250 × 211.63 = 5.29 g.

Why invert the flask 20 times after preparation?

To mix evenly and ensure uniform concentration.

List sources of error when preparing a solution.

Incomplete dissolution, spillage, incorrect volume reading at meniscus.

Define dilution.

Adding solvent to decrease concentration without changing moles of solute.

State the dilution equation.

M₁ V₁ = M₂ V₂.

What remains constant during dilution?

Moles of solute.

Example: What volume of 16.0 M H₂SO₄ to prepare 1.00 L of 2.00 M acid?

V₁ = (2.00 × 1.00) / 16.0 = 0.125 L = 125 mL.

Example: 25 mL of 6.0 M HCl diluted to 500 mL; find M₂.

M₂ = (6.0 × 25) / 500 = 0.30 M.

Explain the molecular view of dilution.

Adding solvent spreads solute particles farther apart, reducing their number per unit volume.

Define solution stoichiometry.

Calculations involving molarities and volumes to determine amounts of reactants or products in solution.

Steps for solving solution stoichiometry problems.

1. Balance equation

2. Convert volumes to moles (n = M × V)

3. Apply mole ratios

4. Convert to desired quantity.

Example: Neutralize 25.0 mL 0.100 M H₂SO₄ with 0.250 M NaOH.

2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O; mol H₂SO₄ = 0.00250; mol NaOH = 0.00500; V = 0.00500 / 0.250 = 0.0200 L = 20.0 mL.

Why is solution stoichiometry important?

Used in titrations and for calculating reactant or product quantities in aqueous reactions.

How does concentration influence reaction rate?

Higher concentration → more frequent collisions → faster rate.

Summarize the relationship between moles, molarity, and volume.

n = M × V links all three and connects equation coefficients to measured volumes.

Key conceptual link between Ch 8 and Ch 9.

Ch 8 relates mass ↔ moles ↔ particles

Ch 9 extends to solutions where volume and molarity replace mass as the measured quantity.

What is the fundamental law connecting all solution reactions?

Conservation of mass and charge applies to dissolved species as well as solids and gases.