Asked by Carrie
                Name and describe two colligative properties that are affected by solute concentration, then list whyone would be affected in that manner.
So this is what i put for my answer
Lowers freezing point: Each additional level of molal concentration lowers the freezing point. The freezing point would be lowered because the concentration cause the molecules to slow down and then freeze
Raises the boiling point: Each additional level of molal concentration raises the boiling point.
I got a point off for this problem and i don't understand why, can i get some help?
            
        So this is what i put for my answer
Lowers freezing point: Each additional level of molal concentration lowers the freezing point. The freezing point would be lowered because the concentration cause the molecules to slow down and then freeze
Raises the boiling point: Each additional level of molal concentration raises the boiling point.
I got a point off for this problem and i don't understand why, can i get some help?
Answers
                    Answered by
            DrBob222
            
    Personally I think the answer is not focused and to the point. It isn't clear what you mean by  "additional level" and the explanation of how the foreign molecules interfere to cause the freezing to be lower doesn't explain anything. I don't think asking a stranger to critique this is what you want; I suggest you go back to the instructor and ask him or her what answer was expected. 
    
                    Answered by
            lala
            
    suppose you wanted to heat a mug of water 250 ml for room temperature 25c to 100c to make a cup of tea. how much energy (in units of calories and joules) would you need form the microwave?
    
                    Answered by
            DrBob222
            
    q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Use 1 cal/g*C for specific heat for the answer in calories; use 4.184 J/g*C for the answer in joules.
    
Use 1 cal/g*C for specific heat for the answer in calories; use 4.184 J/g*C for the answer in joules.
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