Isotonic solution calculations using freezing point and molecular weight MCQs With Answer

Introduction

Isotonic solution calculations using freezing point depression and molecular weight are essential skills for B. Pharm students. This topic links colligative properties, Kf (cryoscopic constant), molality, van’t Hoff factor and practical concepts like isotonicity, E-value and sodium chloride equivalents. Clear understanding of ΔTf = Kf · m · i enables accurate molecular weight determination from experimental freezing point data and precise formulation of parenteral and ophthalmic dosage forms. Mastery of these calculations prevents patient discomfort or harm from hypo- or hypertonic preparations. This set of focused practice items emphasizes stepwise calculation, unit consistency and common approximations (Kf = 1.86 °C·kg·mol−1, isotonic ΔTf ≈ 0.52 °C). Now let’s test your knowledge with 30 MCQs on this topic.

Q1. What defines an isotonic solution in pharmaceutical practice?

• Solution having the same pH as blood
• Solution having the same freezing point as pure water
• Solution having the same osmotic pressure as body fluids
• Solution having the same density as body fluids

Correct Answer: Solution having the same osmotic pressure as body fluids

Q2. What is the typical freezing point depression (ΔTf) used to represent isotonicity with physiological saline?

• 0.10 °C
• 0.52 °C
• 1.86 °C
• 0.90 °C

Correct Answer: 0.52 °C

Q3. What is the cryoscopic constant (Kf) of water commonly used in isotonic calculations?

• 0.52 °C·kg·mol−1
• 1.86 °C·kg·mol−1
• 2.30 °C·kg·mol−1
• 0.92 °C·kg·mol−1

Correct Answer: 1.86 °C·kg·mol−1

Q4. The van’t Hoff factor (i) represents:

• The density change on dissolution
• The number of particles a solute yields in solution
• The pH of an ionic solution
• The boiling point elevation constant

Correct Answer: The number of particles a solute yields in solution

Q5. For a non-electrolyte in water, what is the molality (mol·kg−1) required to produce an isotonic ΔTf of 0.52 °C? (Use Kf = 1.86 °C·kg·mol−1)

• 0.28 mol·kg−1
• 0.52 mol·kg−1
• 1.86 mol·kg−1
• 0.14 mol·kg−1

Correct Answer: 0.28 mol·kg−1

Q6. Using the molality for isotonic non-electrolytes (~0.2796 mol·kg−1), how many moles of solute are needed for 100 mL (≈0.100 kg solvent) to be isotonic?

• 0.0028 mol
• 0.0280 mol
• 0.280 mol
• 0.560 mol

Correct Answer: 0.0280 mol

Q7. How many grams of a non-electrolyte drug with molecular weight 180 g·mol−1 are required per 100 mL of water to make it isotonic? (Use molality ≈0.2796 mol·kg−1)

• 0.50 g
• 1.80 g
• 5.03 g
• 18.00 g

Correct Answer: 5.03 g

Q8. A solute (non-electrolyte) causes a freezing point depression of 0.74 °C when 2.00 g are dissolved in 100 g of water. What is its approximate molecular weight? (Kf = 1.86 °C·kg·mol−1)

• 25.1 g·mol−1
• 50.3 g·mol−1
• 100.6 g·mol−1
• 200.0 g·mol−1

Correct Answer: 50.3 g·mol−1

Q9. Assuming ideal dissociation (i = 2), how many grams of NaCl (MW ≈ 58.44 g·mol−1) are theoretically required per 100 mL to produce isotonicity using the freezing point method?

• 0.18 g
• 0.82 g
• 1.80 g
• 5.03 g

Correct Answer: 0.82 g

Q10. How many milligrams of a non-electrolyte drug of MW 300 g·mol−1 are needed to make 50 mL isotonic? (Use molality ≈0.2796 mol·kg−1)

• 420 mg
• 1,050 mg
• 4,194 mg
• 13,980 mg

Correct Answer: 4,194 mg

Q11. If 0.500 g of a solute lowers the freezing point by 0.31 °C in 50 g of water, what is the solute’s approximate molecular weight? (Kf = 1.86)

• 30 g·mol−1
• 45 g·mol−1
• 60 g·mol−1
• 120 g·mol−1

Correct Answer: 60 g·mol−1

Q12. Which equation correctly expresses freezing point depression for ideal dilute solutions?

• ΔTf = Kf / (m · i)
• ΔTf = Kf × m × i
• ΔTf = m / (Kf × i)
• ΔTf = Kf + m + i

Correct Answer: ΔTf = Kf × m × i

Q13. For a non-electrolyte, dissolving 1.00 mol in 1.00 kg of water will produce which freezing point depression (approx)? (Kf = 1.86)

• 0.52 °C
• 1.00 °C
• 1.86 °C
• 3.72 °C

Correct Answer: 1.86 °C

Q14. For an electrolyte that yields three particles per formula unit (i ≈ 3, e.g., CaCl2), how many grams of CaCl2·2H2O (MW ≈ 147 g·mol−1 for dihydrate) are approximately needed per 100 mL to be isotonic? (Use Kf = 1.86 and isotonic ΔTf = 0.52 °C)

• 0.34 g
• 1.04 g
• 3.40 g
• 10.4 g

Correct Answer: 1.04 g

Q15. Which of the following is NOT a colligative property?

• Freezing point depression
• Vapor pressure lowering
• Boiling point elevation
• Density change

Correct Answer: Density change

Q16. If a drug contributes a ΔTf of 0.20 °C per 100 mL, how much additional ΔTf must be provided by NaCl to reach isotonicity?

• 0.20 °C
• 0.32 °C
• 0.72 °C
• −0.32 °C

Correct Answer: 0.32 °C

Q17. A substance (non-electrolyte) of mass 0.400 g in 200 g water causes ΔTf = 0.12 °C. What is its approximate molecular weight? (Kf = 1.86)

• 15 g·mol−1
• 31 g·mol−1
• 62 g·mol−1
• 124 g·mol−1

Correct Answer: 31 g·mol−1

Q18. The isotonic freezing point depression (≈0.52 °C) corresponds approximately to what physiological osmolarity?

• 30 mOsm·L−1
• 150 mOsm·L−1
• 300 mOsm·L−1
• 900 mOsm·L−1

Correct Answer: 300 mOsm·L−1

Q19. How many grams of sucrose (MW 342 g·mol−1) are required per 100 mL to be isotonic? (Use molality ≈0.2796 mol·kg−1)

• 3.42 g
• 6.84 g
• 9.56 g
• 34.20 g

Correct Answer: 9.56 g

Q20. A drug that dissociates into two ions (i ≈ 2) with MW 150 g·mol−1 is to be made isotonic per 100 mL. Approximately how many grams are required? (Use Kf = 1.86)

• 0.52 g
• 1.05 g
• 2.10 g
• 4.20 g

Correct Answer: 2.10 g

Q21. The E-value of a drug is defined as:

• The grams of drug equivalent to 1 mL of water
• The grams of NaCl that have the same osmotic effect as 1 g of the drug
• The molar concentration required to change freezing point by 1 °C
• The amount of drug that yields a van’t Hoff factor of 1

Correct Answer: The grams of NaCl that have the same osmotic effect as 1 g of the drug

Q22. If the experimentally measured ΔTf for an ionic solute is smaller than the value predicted assuming complete dissociation, this most likely indicates:

• Complete association
• Partial dissociation or ionic interactions
• Higher molecular weight than expected
• Lower Kf than assumed

Correct Answer: Partial dissociation or ionic interactions

Q23. Which solvent has a cryoscopic constant (Kf) close to 1.86 °C·kg·mol−1?

• Benzene
• Water
• Acetone
• Ethanol

Correct Answer: Water

Q24. If dissolving 1.00 g of solute in 100 g water produces ΔTf = 0.93 °C, what is the molecular weight of the solute? (Kf = 1.86)

• 10 g·mol−1
• 20 g·mol−1
• 50 g·mol−1
• 100 g·mol−1

Correct Answer: 20 g·mol−1

Q25. For very dilute aqueous solutions at near 1 atm and 25 °C, the numerical values of molarity and molality are often approximately equal because:

• Water volume changes dramatically on solute addition
• Densities are similar so 1 L ≈ 1 kg for dilute solutions
• Molarity and molality are defined identically
• Concentration units are interchangeable by convention

Correct Answer: Densities are similar so 1 L ≈ 1 kg for dilute solutions

Q26. How many grams of a non-electrolyte drug of MW 165 g·mol−1 are needed to make 250 mL isotonic? (Use molality ≈0.2796 mol·kg−1)

• 1.15 g
• 4.60 g
• 11.53 g
• 46.12 g

Correct Answer: 11.53 g

Q27. The sodium chloride equivalent method (using E-values) is primarily used to:

• Calculate buffer capacity
• Determine sterility of parenterals
• Adjust osmotic properties to achieve isotonicity
• Measure drug purity

Correct Answer: Adjust osmotic properties to achieve isotonicity

Q28. Which information is essential to determine molecular weight from freezing point depression?

• Mass of solute, mass of solvent and Kf
• Boiling point of solvent and pH of solution
• Density of solute and color of solution
• Melting point of pure solute only

Correct Answer: Mass of solute, mass of solvent and Kf

Q29. If a solute produces nearly twice the freezing point depression expected for a single particle solute, this suggests the solute likely:

• Is polymeric with very high MW
• Isomerizes in solution
• Dissociates into two particles (i ≈ 2)
• Has negligible solubility

Correct Answer: Dissociates into two particles (i ≈ 2)

Q30. When using freezing point depression methods for isotonic calculations, which common assumption is made about the solution?

• Non-ideal interactions are always dominant
• The solution behaves ideally and colligative relationships apply
• The solvent Kf changes with solute identity
• The solute always polymerizes on dilution

Correct Answer: The solution behaves ideally and colligative relationships apply

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