Strength of a Solution in g/L: Converting to Molarity and Back

By Aniket Bhardwaj · 11 November 2026 · Physical Chemistry

School laboratory bottles and practical files often give the strength of a solution in grams per litre, while problems and titrations use molarity in moles per litre. Moving between the two is a one-step job, but students lose marks by mixing up grams, moles and litres. This page gives the link and four checked examples.

The relation

Strength (g L⁻¹) = Molarity (mol L⁻¹) × Molar mass (g mol⁻¹)
Molarity = Strength ÷ Molar mass

Strength is the mass of solute (in grams) present in one litre of solution. Some books call it “concentration in g/L”.
Molarity (M) is the amount of solute in moles present in one litre of solution.
Molar mass converts between grams and moles. It is found by adding atomic masses from the periodic table.
Volume must be in litres. If the volume is in mL, divide by 1000 first.

The relation is only a rearrangement of moles = mass ÷ molar mass, divided by volume on both sides. So if you forget it in the exam, you can build it again from that line.

Steps to follow

Write the molar mass of the solute with its unit.
Convert the volume to litres.
If mass and volume are given, first find strength = mass ÷ volume (g L⁻¹).
Divide strength by molar mass to get molarity, or multiply molarity by molar mass to get strength.
Check the answer by a second route, for example by finding moles first.

Worked examples

Example 1: from mass and volume. 4.9 g of H₂SO₄ is dissolved in water to make 500 mL of solution. Find the strength and the molarity.

Molar mass of H₂SO₄ = 2(1.008) + 32.06 + 4(15.999) = 2.016 + 32.06 + 63.996 = 98.07 g mol⁻¹

Volume = 500 mL = 0.500 L

Strength = 4.9 g ÷ 0.500 L = 9.8 g L⁻¹

Molarity = 9.8 ÷ 98.07 = 0.0999 ≈ 0.10 M

Second route: moles = 4.9 ÷ 98.07 = 0.04997 mol; molarity = 0.04997 ÷ 0.500 = 0.0999 M. ✓

Example 2: from molarity to strength. A bottle is labelled 0.25 M NaOH. What is the strength in g L⁻¹?

Molar mass of NaOH = 22.990 + 15.999 + 1.008 = 40.00 g mol⁻¹

Strength = 0.25 × 40.00 = 10 g L⁻¹

So every litre of this solution holds 10 g of NaOH, and 250 mL holds 2.5 g.

Example 3: a given strength of acid. A solution of hydrochloric acid has a strength of 18.25 g L⁻¹. Find its molarity.

Molar mass of HCl = 1.008 + 35.45 = 36.46 g mol⁻¹

Molarity = 18.25 ÷ 36.46 = 0.5006 ≈ 0.50 M

Because 18.25 is almost exactly half of 36.46, the answer is close to 0.50 M, which is a quick sense-check you can do without a calculator.

Example 4: preparing a solution. You must prepare 250 mL of 0.10 M Na₂CO₃. What mass of anhydrous sodium carbonate do you weigh?

Molar mass of Na₂CO₃ = 2(22.990) + 12.011 + 3(15.999) = 45.980 + 12.011 + 47.997 = 105.99 g mol⁻¹

Strength of 0.10 M solution = 0.10 × 105.99 = 10.60 g L⁻¹

Mass for 0.250 L = 10.60 × 0.250 = 2.65 g

Check: 2.65 ÷ 105.99 = 0.0250 mol; 0.0250 ÷ 0.250 L = 0.100 M. ✓

Common mistakes

Using the volume in mL directly. 500 mL must become 0.500 L before dividing.
Dividing when you should multiply. Going from molarity to strength needs multiplication by the molar mass; going the other way needs division.
Using the wrong molar mass, such as the molar mass of NaOH for Na₂CO₃, or forgetting water of crystallisation in a hydrated salt (for Na₂CO₃·10H₂O the molar mass is much larger).
Confusing the volume of water added with the volume of the final solution. Molarity uses the volume of the solution.
Mixing up strength (g L⁻¹) with percentage strength (g per 100 mL).

Quick revision: unit habits that save marks

Most errors in this topic are unit errors, not chemistry errors. Write the unit next to every number as you work. If you divide grams by g mol⁻¹, the grams cancel and you are left with moles; if you then divide by litres, you get mol L⁻¹, which is molarity. When the units on your last line are not mol L⁻¹ for molarity or g L⁻¹ for strength, something went wrong earlier.

Also remember that a solution made by adding 2.65 g of solute to 250 mL of water is slightly different from one made up to the 250 mL mark of a volumetric flask. In the laboratory you dissolve the solid in a little water first and then fill to the mark, so that the final volume of solution is exactly 250 mL. Your calculation assumes this method.

Finally, keep two or three significant figures in the answer, matching the data given. A mass such as 2.65 g is quoted to three figures because the volume and molarity were given to two or three figures.

Where this is used in exams

Topic	How the conversion helps
Practical examinations	Weighing the right mass to prepare a standard solution.
Titration numericals	Many questions give strength; the formula needs molarity.
Some Class 11 solution problems	Switching between g L⁻¹ and mol L⁻¹ before any further step.

Read the exact wording of each question: some ask for “strength”, some for “molarity” and some for “normality”, and they are not the same quantity.

Enter the mass and volume to get the molarity, then compare it with your own working.

Open the Molarity & Concentration Calculator →

Physical Chemistry · CSIR-NET / GATE

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