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Chemistry

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I'm not sure if i did this problem right or not. I keep reading through my book, and i think i followed the equations right, but i'm not confident with my answers.

The problem:

5.0 g of glucose,C6H12O6, is disolved in 500.0 g of acetic acid. What is the new freezing point and boiling point for the solution? Kf acetic acid = 3.90, Kb acetic acid = 3.07 (normal freezing point for acetic acid = 16.60 degrees C, boiling point = 118.5 degrees C)

my work:

# of moles
(5g C6H12O6)(1mol C6H12O6/ 180g C6H12O6)
= 0.028 mol of C6H12O6

molality
(0.028mol/500g CH3COOH)(1000g CH3COOH/1kg)
= 0.056 mol solution

boiling point = 100degreesC+deltaTbp
= 100degreesC+mKbp
= 100degreesC+(0.056m)(3.07 C/m)
= 100.173 degrees C

freezing point = 0 degreesC-deltaTfp
= 0 degreesC-mKfp
= O degreesC-(0.056m)(3.90C/m)
= -0.218 degrees C

now, is that all there is (IF this is correct...)or is there more to solving this equation.

I really need help..Im so confused...Thank you

  • Chemistry - ,

    # of moles
    (5g C6H12O6)(1mol C6H12O6/ 180g C6H12O6)
    = 0.028 mol of C6H12O6

    Your calculations are ok but you didn't finish. After your have delta T you must add to normal boiling point and subtract from normal freezing point. Also, I think you transferred f.p. and b.p. incorrectly from the problem. I don't like to round at each step as I go through a problem; rounding at each step can cause rounding errors at the end. I prefer to just leave those "extra" numbers in the calculator. For example:

    (5.0/180)/0.5 x 3.90 = 0.21667 which rounds to two places as 0.22 (if 5.0 is the correct value from your problem then two s.f. is all you can have). Then -0.22+16.6 = ??


    molality
    (0.028mol/500g CH3COOH)(1000g CH3COOH/1kg)
    = 0.056 mol solution

    boiling point = 100degreesC+deltaTbp
    = 100degreesC+mKbp
    = 100degreesC+(0.056m)(3.07 C/m)
    = 100.173 degrees C
    I think the problem quotes normal boiling point as 118.5

    freezing point = 0 degreesC-deltaTfp
    = 0 degreesC-mKfp
    = O degreesC-(0.056m)(3.90C/m)
    = -0.218 degrees C
    I think the problem quotes normal freezing point as 16.6.

  • Chemistry - ,

    Thank you

    so then would the new points be:

    b.p. = 218.67 C

    f.p. = 16.38 C ?

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