[professordad]

A solution is made containing 7.5 g CH₃OH in 245 g H₂O.  Calculate the molality of CH₃OH.


Ok so what I did was: since molality is mol solute / kg solvent.  We know that mol solute is 7.5 g CH₃OH · (1 mol CH₃OH) / (32.0419 g CH₃OH) or 0.23 mol CH₃OH.  

Then for the kg solvent there's 245 g H₂O which is basically 0.245 kg H₂O.  

So dividing thru we have m = 0.94.

But the right answer is m = 0.271...

[Hunter2]

I  get 0.96 mol/kg

[Borek]

Closer to 0.96 (0.9554) - you shouldn't round down number of moles so early.

But the "correct" answer is definitely incorrect.

[professordad]

Ok thanks for your inputs!

@Borek: I round down early because my book taught me for significant digits calculations -- should I not do so in the future?  Thanks.

[Borek]

You should do the calculations using full precision, and round down only the final result. That is, if you REPORT intermediate result, you should round it down as well, but you should not use rounded value for calculations.