6.853g of sodium cyanide was dissolved in water and brought to volume in a 100ml volumetric flask. The final mass of the solution was found to be 107.166g

My understanding -

A solution is understood when I put a solute into a solvent (water) and dissolves it.

So in this case, if I wanted to find the composition of solution expressed as moles, I must take mass multiply by the molecular weight of Sodium Cyanide, do I use the mass of the solution or the mass of solute ?

With this then I can go on to find molarity, normality as well as weight percent (mass of solute/mass of total solution) and weight/volume percent (mass of solute/volume of total solution).


Mole fraction is the ratio of the number of mole of a component to that of the number of moles of the total mixture. In your case, 6.853g of NaCN equals 6.853/49.002 = 0.139 moles.

The rest is water i.e. 107.166 - 6.853 = 100.313g. this corresponds to (100.313/18.015) = 5.568 moles.

Now, total moles = 0.1398 + 5.568 = 5.708moles.

The mole fraction of NaCN = 0.139/5.708 = 0.024.

  • $\begingroup$ If we do it like this, how do we calculate composition of solution expressed as molarity / normality ? For example, molarity, 6.853g of NaCN has 0.139 moles. I do not know the volume of NaCN in the solution. $\endgroup$ – user69734 Nov 2 '18 at 6:59
  • $\begingroup$ And the same problem for normality. Thanks ... $\endgroup$ – user69734 Nov 2 '18 at 7:00
  • $\begingroup$ Use proper significant figures $6.853/49.002 = 0.1399$ moles. (100.313/18.015) = 5.5683 moles $\endgroup$ – MaxW Nov 2 '18 at 14:31
  • $\begingroup$ @user69734 - Molarity = moles/liter = (0.1399 moles)/(0.1 liter) = 1.399 molar. // A bit undefined as to how exact the 100 ml volumetric flask is. But I'd expect at least 0.100 and but not quite 0.1000. $\endgroup$ – MaxW Nov 2 '18 at 14:41
  • $\begingroup$ @user69734 - "% by weight" has nothing to do with the "volume" of the solute, in this case NaCN. // A "5% w/w" means weight of solute divided by weight of solvent. So 5 grams of solute per 100 grams of solvent. // A "5%" solution would be 5 grams of solute per 100 grams of solution (solvent+solute). $\endgroup$ – MaxW Nov 2 '18 at 14:47

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