CCommons
Search✍️ WriteSign in
🧪 NEET Chemistry❓ Question

Converting molarity to molality using density: I keep getting 1.67 m instead of 2.38 m. Where is my mistake?

Zoya Khan12th Sample

12 Sept 2026 · 1 min read

I have been staring at this one for almost an hour and I would like to understand it properly rather than memorise a formula.

The problem: a 2 M aqueous solution of glucose (M=180M = 180 g/mol) has a density of 1.2 g/mL. Find its molality.

What I did: 2 mol per litre of solution. Mass of 1 L of solution is 1200 g = 1.2 kg. Molality = 2/1.2 = 1.67 m.

The answer given in the book is 2.38 m. I can see the numbers are not very different but the difference is too big to ignore. I suspect I'm dividing by the wrong mass, but I'm not sure which mass is the right one, and why.

I would also like to know if there is a general conversion formula between molarity, molality, mass percent and density, because the NCERT exercises have several of those and I don't want to redo the same reasoning each time. And what is ppm and when is it used (the chapter mentions it)?

If someone can write the solution in a way that I can reuse for any similar question, that would be wonderful.

WhatsAppPost on X

6 answers & replies

Sign in to join the conversation.

  • [+]Siddharth RaoDropperSample· 1mo ago

    Pick a basis and write out the whole table. This always works. Take 1 L of solution.

    • Mass of solution = 1200 g
    • Moles of glucose = 2, so mass of glucose = 2 × 180 = 360 g
    • Mass of water (solvent) = 1200 − 360 = 840 g = 0.840 kg
    • Molality = 2/0.840 = 2.38 m ✓.

    You used 1.2 kg, the mass of the whole solution. Molality needs the mass of the solvent only. The same table gives mole fraction too: water = 840/18 = 46.7 mol, so xglucose=2/48.7≈0.041x_{glucose} = 2/48.7 \approx 0.041.

    • [+]Zoya Khan12thSample· 1mo agoAuthor

      I think I see it. The basis method is going to be my tool from now on. I divided by the whole solution because 'kg' was sitting right there as a number. Thank you everyone, truly.

  • [+]Harsh Gupta12thSample· 1mo ago

    General formula I use: M=10 d wM0M = \dfrac{10\,d\,w}{M_0} where dd is the density in g/L and ww the mass %. Just plug in 1200 for d here and you get the molarity directly, cool right? Then molality comes by dividing... I forget, ask someone smarter 😂

    • [+]Ishita GoyalDropperSample· 1mo ago

      Harsh, if dd is in g/L the 10 doesn't belong. With dd in g/mL (that's how it's given in most problems) and ww as mass percent:

      M=10 d wM0,m=1000 wM0 (100−w)M = \frac{10\, d\, w}{M_0}, \qquad m = \frac{1000\, w}{M_0\,(100 - w)}

      Derivation of the first one: 100 g of solution contains ww g of solute, so moles =w/M0= w/M_0; volume of that 100 g in litres =100/(1000 d)= 100/(1000\,d). Divide: w/M0100/(1000d)=10dwM0\frac{w/M_0}{100/(1000d)} = \frac{10dw}{M_0}. Put the density in g/L and you inflate it by 1000.

  • [+]Mehak AroraDropperSample· 1mo ago

    A worked one using the formulas: conc. sulphuric acid is 98% by mass with density 1.84 g/mL (M0=98M_0 = 98).

    M=10×1.84×9898=18.4 MM = \frac{10\times1.84\times98}{98} = 18.4\ \text{M}

    m=1000×9898×2=500 mm = \frac{1000\times98}{98\times 2} = 500\ m

    Here molarity is nowhere near molality. That is the point: when the solution is dense or concentrated they are far apart, so you can't use M for m.

  • [+]Rahul DubeyDropperSample· 1mo ago

    ppm (parts per million) is for very dilute things, like the amount of a pollutant in water or oxygen dissolved in water:

    ppm=mass of solutetotal mass of solution×106\text{ppm} = \frac{\text{mass of solute}}{\text{total mass of solution}}\times10^6

    It's also possible to do volume-based ppm for gases. NCERT mentions it briefly, not a big scoring topic, but it's nice when units are tiny.

More from Zoya Khan

Keep reading