What we are making and the tools we need
Molarity (M, mol/L) is the number of moles of solute in 1 litre of solution. A standard solution has a known, accurate molarity.
Tools:
- Analytical balance: weighs to 0.01 g or better.
- Beaker and glass rod: to dissolve the solute.
- Funnel: to pour without spilling.
- Volumetric flask: a flat-bottomed flask with a long thin neck and one ring mark. It is made to hold one exact volume (for example 250.0 mL) at 20 °C when the liquid touches the mark.
- Wash bottle and dropper: to add water drop by drop.
A measuring cylinder is not accurate enough, and a conical flask has no exact volume mark.
Step 1: calculate the mass to weigh
Use two short formulas:
moles n = molarity c × volume V (in litres)
mass m = n × molar mass M
Example: 250 mL of 0.1 mol/L NaCl. V = 250 ÷ 1000 = 0.25 L. n = 0.1 × 0.25 = 0.025 mol. M(NaCl) = 23 + 35.5 = 58.5 g/mol. m = 0.025 × 58.5 = 1.46 g.
Always convert mL to L first. That one step is where many marks are lost.
Step 2: weigh, dissolve, transfer and rinse
- Weigh the required mass in a clean dry beaker. Note the real mass you weighed (it may be 1.47 g, not exactly 1.46 g; use the real value to find the real molarity).
- Add a small amount of distilled water (about 50 mL) and stir with the glass rod until it all dissolves.
- Place a funnel in the neck of the volumetric flask. Pour the solution in, guiding it down the glass rod.
- Rinse the beaker, the rod and the funnel with a little distilled water three times, pouring every rinse into the flask. This makes sure all the solute reaches the flask.
Step 3: fill to the mark and mix
Add distilled water until the level is about 1 cm below the ring mark. Remove the funnel. Now add water drop by drop with a dropper until the bottom of the curved surface (the meniscus) touches the ring mark. Keep your eye level with the mark, not above or below it.
Put on the stopper, hold it with your finger, and turn the flask upside down and back about 10 times so the solution mixes fully. Label it with the name, molarity, date and your name.
Error analysis: what can go wrong
| Mistake | What happens | Effect on molarity |
|---|---|---|
| Going past the ring mark | Volume is more than 250 mL | Too low (diluted) |
| Not rinsing the beaker or funnel | Some solute is lost | Too low |
| Eye above or below the mark | Volume misjudged (parallax) | Slightly too high or too low |
| Weighing wet or damp solute | Real solute mass is less | Too low |
| Using a hot solution | Liquid expands, so there will be less when cool | Too high after cooling |
| Spilling while transferring | Solute is lost | Too low |
| Flask rinsed with water before use | No effect: extra water is ok because you fill to the mark anyway | None |
A flask rinsed with a different solution is wrong. A flask that is wet with distilled water is fine.
Size of the error
If you overshoot a 250 mL flask by 5 mL, the real volume is 255 mL, so real molarity = 0.025 ÷ 0.255 = 0.098 mol/L, about 2% low. Small error, but it matters in careful work.
Try it: predict, then check
Before step 6, predict the mass for 100 mL of 0.5 mol/L NaCl. Then check with the slider. At home, mix 1 level teaspoon of salt in a glass of water and then in a big jug. The same salt makes a stronger solution in the small glass: concentration changes with the volume. Count the "drop by drop" step as a practice game: how close to a line can you fill a bottle by dropper?
Key formulas and definitions
- n = c × V (V in litres)
- m = n × M = c × V × M
- c (real) = moles of solute ÷ actual volume of solution in litres
- 1 L = 1000 mL; 250 mL = 0.25 L
- Fill to the mark with the bottom of the meniscus on the ring and the eye level with the ring
Worked examples
1. Find the mass of NaCl needed to make 250 mL of 0.1 mol/L solution (M = 58.5 g/mol).
V = 0.25 L. n = 0.1 × 0.25 = 0.025 mol. m = 0.025 × 58.5 = 1.46 g.
2. How many grams of NaCl are needed for 500 mL of 0.2 mol/L solution?
V = 0.5 L. n = 0.2 × 0.5 = 0.1 mol. m = 0.1 × 58.5 = 5.85 g.
3. What mass of NaOH (M = 40 g/mol) is needed for 100 mL of 0.5 mol/L solution?
V = 0.1 L. n = 0.5 × 0.1 = 0.05 mol. m = 0.05 × 40 = 2.0 g.
4. Find the mass of Na₂CO₃ (M = 106 g/mol) for 250 mL of 0.1 mol/L solution.
n = 0.1 × 0.25 = 0.025 mol. m = 0.025 × 106 = 2.65 g.
5. A student dissolves 1.17 g NaCl (M = 58.5 g/mol) and makes it up to 200 mL. Find the molarity.
n = 1.17 ÷ 58.5 = 0.02 mol. V = 0.2 L. c = 0.02 ÷ 0.2 = 0.1 mol/L.
6. A student aimed for 0.1 mol/L in a 250 mL flask but added 5 mL too much water. What is the real molarity?
Moles = 0.1 × 0.25 = 0.025 mol. Real volume = 255 mL = 0.255 L. c = 0.025 ÷ 0.255 = 0.098 mol/L (about 2% low).
Common mistakes
- Using volume in mL in the formula. Always change to litres first (250 mL = 0.25 L).
- Putting the solid straight into the volumetric flask. Dissolve in a beaker first, then transfer.
- Filling to the mark with the eye above or below it, or reading the top of the curve instead of the bottom.
- Forgetting to rinse the beaker, rod and funnel into the flask, so some solute is lost.