Question

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

Answer 1

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

<b> 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 = ??</b>

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
<b>I think the problem quotes normal boiling point as 118.5</b>

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

Answer 2

Thank you

so then would the new points be:

b.p. = 218.67 C

f.p. = 16.38 C ?