1.1 Calculate the concentration of the unknown.
Reveal worked answer
x = (0.455 − 0.018)/0.842 = 0.519 in the concentration unit used for the standards.
Original practice · not an official paper
Paper 1B skills practice
Six independent scenarios, 18 short tasks. Work first, then reveal the answer and compare your reasoning.
These are original educational exercises, not reproduced or predicted IB examination questions.
0 of 18 tasks reviewed.
Calibration curve
A student prepares a set of standards and records an instrument response. A linear fit is y = 0.842x + 0.018 with R² = 0.997. An unknown gives y = 0.455.
1.2 What does R² = 0.997 tell you, and what does it not prove?
Reveal worked answer
It indicates that the chosen linear trend explains almost all variation in the calibration data. It does not prove the method is unbiased, chemically valid, or free from systematic error.
1.3 Why is an unknown outside the standards’ concentration range less secure?
Reveal worked answer
That would require extrapolation. The linear relationship has not been experimentally established beyond the calibrated range, so model error can grow.
Useful tools: Calibration curve · Percentage uncertainty · Significant figures
Titration data
A 25.00 cm³ sample is titrated three times. Concordant titres are 24.62 cm³ and 24.58 cm³. The reacting stoichiometry is 1:1 and the titrant concentration is 0.1000 mol dm⁻³.
2.1 Calculate the mean concordant titre.
Reveal worked answer
(24.62 + 24.58)/2 = 24.60 cm³.
2.2 Calculate the amount of titrant at the mean titre.
Reveal worked answer
n = CV = 0.1000 × 0.02460 = 2.460 × 10⁻³ mol.
2.3 If each burette reading contributes ±0.05 cm³, explain why the titre uncertainty is not simply ±0.05 cm³.
Reveal worked answer
A titre is the difference between two readings. Under the course linear rule, the absolute reading uncertainties add, so the uncertainty associated with the difference is larger.
Useful tools: Titration stoichiometry · Percentage uncertainty
Calorimetry
A reaction warms 100.0 g of a water-like solution from 20.0 °C to 27.5 °C. Use c = 4.18 J g⁻¹ K⁻¹. The reaction amount represented is 0.0500 mol.
3.1 Calculate the heat gained by the solution.
Reveal worked answer
q = mcΔT = 100.0 × 4.18 × 7.5 = 3135 J = 3.135 kJ.
3.2 State the sign of the reaction enthalpy under the simple heat-exchange model.
Reveal worked answer
The solution gains heat, so the reaction loses heat: qreaction is negative and the calculated ΔH is negative.
3.3 Give one reason the magnitude could be underestimated in a simple cup experiment.
Reveal worked answer
Heat can be transferred to the surroundings or apparatus rather than being included in the measured solution temperature rise.
Useful tools: Calorimetry · Error & difference
Reaction-rate comparison
The same visible endpoint is used in four clock-style trials. One trial reaches the endpoint in 40.0 s and another in 20.0 s.
4.1 Calculate the 1/t rate proxies for the two trials.
Reveal worked answer
1/40.0 = 0.0250 s⁻¹ and 1/20.0 = 0.0500 s⁻¹.
4.2 What condition is needed before comparing 1/t as a rate proxy?
Reveal worked answer
The same observable endpoint should correspond to the same extent/threshold in each trial, with the comparison otherwise meaningful.
4.3 If the relevant concentration doubled and 1/t doubled while other factors stayed constant, what order is suggested?
Reveal worked answer
First order with respect to that reactant because the rate proxy changes by the same factor as concentration.
Useful tools: Reaction rate · Rate order (HL)
Gas data
A gas sample has P = 101.3 kPa, V = 2.40 dm³ and T = 298.15 K. Assume ideal behaviour.
5.1 Calculate the amount of gas.
Reveal worked answer
n = PV/RT = (101.3 × 2.40)/(8.314 × 298.15) ≈ 0.0981 mol.
5.2 Why must temperature be converted to kelvin?
Reveal worked answer
The ideal-gas equation uses absolute temperature; the proportionality between thermal state and T requires the zero of the Kelvin scale.
5.3 State one reason real-gas behaviour may deviate from the ideal model.
Reveal worked answer
Real particles have intermolecular interactions and finite molecular volume; either can matter under some conditions.
Useful tools: Ideal gas equation · Moles, mass & particles
Uncertainty and significant figures
A volumetric measurement is reported as 25.00 ± 0.05 cm³ and a second independent measured factor is 0.1000 ± 0.0005 mol dm⁻³.
6.1 Calculate the percentage uncertainty in the volume.
Reveal worked answer
(0.05/25.00) × 100 = 0.20%.
6.2 For a product of these two measured quantities, what course-level propagation rule is used?
Reveal worked answer
Add the percentage uncertainties of the multiplied measured quantities.
6.3 Why should intermediate calculator values normally retain extra digits?
Reveal worked answer
Premature rounding can accumulate error. Keep guard digits in working and round the final reported value to a justified precision.
Useful tools: Percentage uncertainty · Uncertainty propagation · Significant figures