Common Registration Assessment, Part 1

Isotonicity by freezing point depression

Three original questions on adjusting solutions to isotonicity, with the working shown and the usual traps pointed out.

Published 4 October 2026. Independent revision material, not GPhC questions, not for patient care.

Vintage laboratory glassware with a graduated cylinder
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A solution is isotonic with body fluids when it freezes at minus 0.52 degrees Celsius. Each ingredient lowers the freezing point a little, and sodium chloride is added to make up the shortfall.

The questions here use fictional drugs and practise only the arithmetic. A 1% w/v sodium chloride solution lowers the freezing point by 0.576 degrees, which is the anchor for every question of this type.

The method

  1. Write the target depression, 0.52 degrees.
  2. Work out the freezing point contribution of every ingredient at its final concentration.
  3. Subtract the total contribution from 0.52 to find the shortfall.
  4. Divide the shortfall by 0.576 to get the sodium chloride needed in % w/v.
  5. Convert the percentage to a mass for the volume in the question.

Three practice questions

Work each one on paper first, then open the answer. All drugs and patients are fictional.

Question 1

A 1% w/v solution of a fictional drug lowers the freezing point by 0.12 degrees. How much sodium chloride, in grams, makes 100 mL of a 1% solution isotonic?

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Answer: 0.69 g

Working: Shortfall = 0.52 - 0.12 = 0.40. NaCl % = 0.40 / 0.576 = 0.694% w/v. For 100 mL that is 0.694 g, about 0.69 g.

Question 2

For the same fictional drug, a 2% w/v solution lowers the freezing point by 0.16 degrees. How much sodium chloride is needed for 50 mL?

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Answer: 0.31 g

Working: Shortfall = 0.52 - 0.16 = 0.36. NaCl % = 0.36 / 0.576 = 0.625% w/v. For 50 mL: 0.625 / 2 = 0.3125 g, about 0.31 g.

Question 3

A fictional eye drop contains drug X 1% (depression 0.1 degrees) and a preservative contributing 0.06 degrees. How much sodium chloride makes 200 mL isotonic?

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Answer: 1.25 g

Working: Total contribution = 0.1 + 0.06 = 0.16. Shortfall = 0.52 - 0.16 = 0.36. NaCl % = 0.36 / 0.576 = 0.625% w/v. For 200 mL: 2 x 0.625 = 1.25 g.

Anchor on 0.576

Every freezing point question is one shortfall divided by 0.576. Write the anchor down first, then line up the contributions underneath it.

Where marks are lost

  • Forgetting to scale a 1% contribution up to the actual concentration in the product.
  • Adding contributions instead of subtracting the total from 0.52.
  • Stopping at the percentage and not converting to a mass for the stated volume.

Frequently asked questions

Why 0.52 degrees?

Body fluids freeze at about minus 0.52 degrees Celsius, so an isotonic solution must show the same depression. The exam provides this figure.

Is the sodium chloride figure exact?

It is as exact as the data given. In practice you would weigh to the precision of your balance, and in the exam you round as the question directs.

Sources

The questions above are original and use fictional drugs. PreRegExamPrep is not affiliated with or endorsed by the General Pharmaceutical Council.

Practise until the method is automatic

Try 15 free questions with worked answers. No sign-up required.

More calculation topics

See also the formula sheet, the eight sample questions and the approved calculator page.