Solutions & Concentration • Step-by-step solutions
$$M = \frac{\text{moles of solute}}{\text{liters of solution}}$$
The molarity formula calculates the concentration of a solution in moles per liter (mol/L). This fundamental concept in chemistry is essential for stoichiometry, titrations, and solution preparation. Molarity (M) represents the number of moles of solute dissolved in one liter of solution.
Key relationships:
This formula is crucial for laboratory work, pharmaceutical preparations, and industrial processes. Understanding molarity allows chemists to control reaction stoichiometry, prepare standard solutions, and perform quantitative analyses.
Where n = moles of solute, V = liters of solution
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Enter parameters to see solution steps.
Basic definition:
$$M = \frac{\text{moles of solute}}{\text{liters of solution}}$$
Calculating moles from mass:
$$\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$$
Dilution equation:
$$M_1V_1 = M_2V_2$$
Concentration conversions:
$$\text{PPM} = \frac{\text{mg solute}}{\text{L solution}}$$
Molarity is a measure of the concentration of a solution, defined as the number of moles of solute per liter of solution. It is one of the most common ways to express concentration in chemistry and is essential for stoichiometric calculations, titrations, and preparing standard solutions. The unit of molarity is moles per liter (mol/L) or simply M.
The complete mathematical formulation of the molarity formula:
Where:
Key applications of molarity calculations:
Key concepts in solution chemistry:
\(M = n/V\)
\(n = m/MM\), \(M_1V_1 = M_2V_2\), \(PPM = mg/L\)
Where n = moles, m = mass, MM = molar mass.
Accurate weighing → Dissolution → Volume adjustment → Mixing
How many moles of NaCl are contained in 2.0 L of a 0.50 M NaCl solution?
Using the molarity formula: M = n/V
Rearranging to solve for moles: n = M × V
Given: M = 0.50 M, V = 2.0 L
Step 1: n = 0.50 mol/L × 2.0 L
Step 2: n = 1.0 mol
The answer is C) 1.0 mol.
This problem demonstrates the direct application of the molarity formula. Molarity is the ratio of moles of solute to liters of solution. When you know the molarity and volume, you can calculate the moles of solute by multiplying these values. This is a fundamental calculation in chemistry laboratories and is essential for preparing solutions of known concentration.
Molarity: Moles of solute per liter of solution
Mole: Amount of substance containing Avogadro's number of particles
Solute: Substance being dissolved in a solution
• M = n/V (molarity equation)
• n = M × V (rearranged for moles)
• Always use liters for volume
• Remember: Molarity × Volume = Moles
• Check units: mol/L × L = mol
• Use dimensional analysis to verify
• Using mL instead of L for volume
• Confusing molarity with molality
• Not checking unit consistency
How would you prepare 500 mL of a 0.25 M solution of potassium chloride (KCl)? Show all calculations and explain the procedure step by step.
Step 1: Calculate moles needed using n = M × V
Step 2: V = 500 mL = 0.500 L
Step 3: n = 0.25 mol/L × 0.500 L = 0.125 mol
Step 4: Calculate mass needed using m = n × MM
Step 5: MM of KCl = 39.10 + 35.45 = 74.55 g/mol
Step 6: m = 0.125 mol × 74.55 g/mol = 9.32 g
Procedure: Weigh 9.32 g KCl, dissolve in ~400 mL water, transfer to 500 mL volumetric flask, and dilute to mark.
This problem combines molarity calculations with solution preparation techniques. First, calculate the moles of solute needed using the molarity formula. Then convert moles to mass using the molar mass. Finally, follow proper laboratory procedure: weigh the calculated mass, dissolve in a smaller volume of solvent, transfer to a volumetric flask, and dilute to the calibration mark. This ensures accurate concentration.
Volumetric Flask: Precise glassware for preparing solutions
Molar Mass: Mass of one mole of substance
Avogadro's Number: 6.022×10²³ particles per mole
• Use volumetric flasks for accuracy
• Dissolve before diluting to mark
• Mix thoroughly after dilution
• Always calculate mass before weighing
• Use analytical balance for precision
• Swirl to mix, don't shake vigorously
• Adding solute directly to final volume
• Not accounting for dissolution volume
• Using wrong glassware for preparation
You have 100 mL of a 2.0 M stock solution of sodium hydroxide (NaOH). You want to prepare 500 mL of a 0.40 M solution. Will you have enough stock solution? If yes, how much of the stock solution should you use?
Step 1: Use dilution equation: M₁V₁ = M₂V₂
Step 2: Given: M₁ = 2.0 M, M₂ = 0.40 M, V₂ = 500 mL
Step 3: 2.0 M × V₁ = 0.40 M × 500 mL
Step 4: V₁ = (0.40 M × 500 mL) / 2.0 M
Step 5: V₁ = 200 mL / 2.0 = 100 mL
Yes, you have exactly enough stock solution. Use 100 mL of the 2.0 M stock solution and dilute to 500 mL.
The dilution equation (M₁V₁ = M₂V₂) is based on the principle that the number of moles of solute remains constant during dilution. Only the amount of solvent changes. This problem demonstrates how to calculate the volume of concentrated solution needed to prepare a diluted solution of known concentration. The result shows that exactly all of the available stock solution is needed.
Stock Solution: Concentrated solution used for dilution
Dilution: Adding solvent to decrease concentration
Concentration: Amount of solute per unit volume
• Moles remain constant during dilution
• M₁V₁ = M₂V₂ (dilution equation)
• More dilute = less concentrated
• Always use the dilution equation
• Check: concentrated volume < dilute volume
• Verify: concentrated M > dilute M
• Forgetting to keep moles constant
• Using addition instead of multiplication
• Not checking if stock solution is sufficient
In a titration experiment, 25.0 mL of HCl solution of unknown concentration is neutralized by 32.4 mL of 0.150 M NaOH. The balanced equation is: HCl + NaOH → NaCl + H₂O. Calculate the molarity of the HCl solution.
Step 1: Calculate moles of NaOH used: n = M × V
Step 2: n_NaOH = 0.150 mol/L × 0.0324 L = 0.00486 mol
Step 3: From balanced equation, mole ratio is 1:1
Step 4: n_HCl = n_NaOH = 0.00486 mol
Step 5: Calculate molarity of HCl: M = n/V
Step 6: M_HCl = 0.00486 mol / 0.0250 L = 0.194 M
The molarity of the HCl solution is 0.194 M.
This problem demonstrates how molarity is used in stoichiometric calculations. First, calculate the moles of the known solution (NaOH). Then use the balanced chemical equation to determine the mole ratio between reactants. Finally, calculate the molarity of the unknown solution using the mole amount and volume. This is the basis for analytical chemistry techniques like acid-base titrations.
Titration: Technique for determining concentration
Equivalence Point: When stoichiometric amounts react
Stoichiometry: Quantitative relationships in reactions
• Balanced equation shows mole ratios
• n = M × V for solution calculations
• At equivalence: moles acid = moles base
• Always balance the equation first
• Convert volumes to liters
• Use mole ratios from balanced equation
• Not balancing the chemical equation
• Using incorrect mole ratios
• Forgetting to convert mL to L
What is the concentration of chloride ions in a 0.20 M solution of aluminum chloride (AlCl₃)?
Step 1: Write the dissociation equation: AlCl₃ → Al³⁺ + 3Cl⁻
Step 2: From the equation, 1 mole of AlCl₃ produces 3 moles of Cl⁻
Step 3: If [AlCl₃] = 0.20 M, then [Cl⁻] = 3 × 0.20 M
Step 4: [Cl⁻] = 0.60 M
The answer is C) 0.60 M.
This problem addresses the concept of ion dissociation in solution. When ionic compounds dissolve, they separate into their constituent ions. Aluminum chloride dissociates into one aluminum ion and three chloride ions. Therefore, the concentration of chloride ions is three times the concentration of the aluminum chloride compound. This is important for understanding electrolyte solutions and colligative properties.
Electrolyte: Substance that conducts electricity in solution
Dissociation: Separation into ions in solution
Ionic Compound: Made of charged particles
• Ionic compounds dissociate in solution
• Coefficients become ion concentrations
• Charge balance must be maintained
• Write the dissociation equation first
• Count the number of each ion
• Multiply by coefficient for concentration
• Forgetting to consider dissociation
• Not counting all ions properly
• Confusing compound with ion concentration
Q: What's the difference between molarity and molality?
A: Molarity (M) is moles of solute per liter of solution, while molality (m) is moles of solute per kilogram of solvent. The key difference is that molarity uses the volume of the solution, while molality uses the mass of the solvent.
Molarity changes with temperature because volume changes with temperature. Molality is temperature-independent because mass doesn't change with temperature. Molality is preferred for colligative property calculations (boiling point elevation, freezing point depression) because these properties depend on the number of particles per mass of solvent.
Q: How does temperature affect molarity calculations?
A: Temperature affects molarity because the volume of a solution changes with temperature. As temperature increases, liquids expand, so the same amount of solute occupies a larger volume, decreasing the molarity. This is why molarity is temperature-dependent.
For precise analytical work, it's important to specify the temperature at which molarity was determined. For processes where temperature changes significantly, molality (moles per kg solvent) is often preferred as it doesn't change with temperature. The relationship is: V_T = V_ref[1 + α(T - T_ref)], where α is the thermal expansion coefficient.
Q: What's the relationship between molarity and normality?
A: Normality (N) is related to molarity (M) by the equation: N = M × n, where n is the number of equivalents per mole of substance. For acids, n is the number of H⁺ ions that can be donated; for bases, it's the number of OH⁻ ions that can be donated; for redox reactions, it's the number of electrons transferred.
For example, 1 M HCl = 1 N HCl (one H⁺), but 1 M H₂SO₄ = 2 N H₂SO₄ (two H⁺). Normality is less commonly used today as it's reaction-specific, whereas molarity is a more general concentration unit. However, it's still used in some titration contexts.