David--I alerted Bob Pursley to this problem and he figured out the problem. Smart man Bob Pursley.
The problem says "1.00 kg of 1.48 m CH3OH ....." That means there is a kg of the SOLUTION, not 1 kg of the solvent. Therefore, you have fewer than 1.48 moles in the kg of SOLUTION.
How many grams of water would you add to 1.00 kg of 1.48 m CH3OH(aq) to reduce the molality to 1.00 m CH3OH. Use two sig figs.
I get 4.8 * 10^2 grams and its not right. I want to put this up here again to see if there are any errors that could have possibly been made
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so would the answer be 4.8 * 10 ^ -2?? if its fewer?
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