G3 Chemistry tuition can help your child convert cubic centimetres to cubic decimetres by connecting the volume unit to the concentration unit before calculating. When concentration is in mol/dm³, use volume in dm³ in n = cV. Divide a volume in cm³ by 1000 to express it in dm³.

For a Punggol parent, a useful first question is: ‘If one cubic decimetre contains 1000 cubic centimetres, is 25 cm³ more or less than one dm³?’ It is less: 25 cm³ = 0.025 dm³. This size check can catch a factor-of-1000 error before the calculator produces a convincing-looking answer.

The 2027 SEC G3 Chemistry syllabus K324 includes solution concentration in mol/dm³ and g/dm³ and its use in volumetric calculations. This guide isolates the unit-conversion step. Its examples are paper exercises, and laboratory work should remain under the school’s supervision.

Confirm the student’s actual subject level and examination year. PG1, PG2 and PG3 are Posting Groups; they do not replace checking each subject’s G level. Ask a provider about subject availability before arranging specialist support.

A concentration of 0.200 mol/dm³ means 0.200 mol of solute per cubic decimetre of solution. It does not mean every sample contains 0.200 mol. The amount depends on how much solution is taken.

Write the known quantities with their units before substituting: c = 0.200 mol/dm³ and V = 25.0 cm³. The volume unit does not yet match the denominator of the concentration unit.

Convert V first. A decimetre is 10 centimetres, so a cube one decimetre on each side contains 10 × 10 × 10 = 1000 cm³. This is why the volume conversion uses 1000 rather than the length-conversion factor of 10.

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For 25.0 cm³ of a 0.200 mol/dm³ solution, V = 25.0/1000 = 0.0250 dm³. Then n = cV = 0.200 × 0.0250 = 0.00500 mol.

Check the scale: the sample is one-fortieth of a cubic decimetre, so it should contain one-fortieth of 0.200 mol. That is 0.00500 mol, agreeing with the calculation.

If the child obtains 5.00 mol, ask which unit was used for V. Multiplying 0.200 directly by 25.0 treats the sample as 25.0 dm³ rather than 25.0 cm³. The formula is not the problem; the substitution is.

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When c is in g/dm³, multiplying by volume in dm³ gives a mass in grams. For example, 20.0 g/dm³ × 0.0250 dm³ = 0.500 g. The units identify what the result means.

To convert that mass into amount in moles, divide by the appropriate molar mass in g/mol. If the stated solute has molar mass 40.0 g/mol, n = 0.500/40.0 = 0.0125 mol.

Do not label cV as moles automatically. Read whether the question provides g/dm³ or mol/dm³. A tutor can diagnose this by asking the student to predict the answer unit before calculating.

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Suppose 25.0 cm³ of 0.200 mol/dm³ hydrochloric acid reacts completely with sodium hydroxide. The balanced equation HCl + NaOH → NaCl + H₂O gives a 1:1 mole ratio. The amount of NaOH reacting is therefore 0.00500 mol.

If that amount is contained in 20.0 cm³ of NaOH solution, its concentration is 0.00500/0.0200 = 0.250 mol/dm³. Convert the second volume separately; converting the acid volume does not automatically convert the alkali volume.

Use the balanced equation for each new reaction. Equal reacting volumes do not imply equal amounts unless the concentrations and reaction conditions support that conclusion. Other reactions can have different mole ratios.

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Try this independently: find the amount in 40.0 cm³ of a 0.150 mol/dm³ solution. Convert to 0.0400 dm³, then calculate 0.150 × 0.0400 = 0.00600 mol.

Now reverse it: 0.0120 mol is dissolved to make 60.0 cm³ of solution. The concentration is 0.0120/0.0600 = 0.200 mol/dm³. The volume is the final solution volume, not automatically the volume of water initially added.

Continue with the existing chemical-calculations guide. A useful correction routine is quantity, unit, conversion, equation, calculation and size check.

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Use the Primary, PSLE and SEC subject directory to find related questions. Bring one recent piece of work and the teacher’s feedback to a consultation. Ask which step needs repair and how the student will demonstrate progress on a new task.

The Clementi Secondary 1 Mathematics guide illustrates diagnosis, sequenced practice and focused 3-pax teaching. Use these principles when asking how support will fit your child’s actual subject and starting point.

Official reference: SEAB 2027 SEC G3 Chemistry syllabus K324, checked 12 October 2026. The exercises here are original teaching examples, not SEAB examination questions or official model answers.