DP Chemistry (first assessment 2025)

Question 19M.2.SL.TZ2.5b(ii)

Date	May 2019	Marks available	[Maximum mark: 2]	Reference code	19M.2.SL.TZ2.5b(ii)
Level	SL	Paper	2	Time zone	TZ2
Command term	Calculate	Question number	b(ii)	Adapted from	N/A

b(ii).

[Maximum mark: 2]

19M.2.SL.TZ2.5b(ii)

(b(ii))

100.0 cm³ of soda water contains 3.0 × 10⁻² g NaHCO₃.

Calculate the concentration of NaHCO₃ in mol dm⁻³.

[2]

Markscheme

«molar mass of NaHCO₃ =» 84.01 «g mol^–1» [✔]

«concentration = $\frac{{3.0 \times {{10}^{ - 2}}{\text{ g}}}}{{84.01{\text{ g mo}}{{\text{l}}^{ - 1}}}} \times \frac{1}{{0.100{\text{ d}}{{\text{m}}^3}}} =$ » 3.6 × 10^–3 «mol dm^–3» [✔]

Note: Award [2] for correct final answer.

Examiners report

Very well answered. Most candidates calculated the molar concentration correctly.

Syllabus sections

Structure 1. Models of the particulate nature of matter » Structure 1.4—Counting particles by mass: The mole » Structure 1.4.5—The molar concentration is determined by the amount of solute and the volume of solution. Solve problems involving the molar concentration, amount of solute and volume of solution.

Question 19M.2.SL.TZ2.5b(ii)

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