Question #e16a6
1 Answer
Here's what I got.
Explanation:
The key here is the difference between the ebullioscopic constant,
You know that you have--see here for the source
#K_b = 0.512^@"C kg mol"^(-1)# #K_f = 1.86^@"C kg mol"^(-1)#
Now, the boiling-point elevation is calculated using
#DeltaT_"b" = i * K_b * b#
Here
#i# is the van't Hoff factor#b# is the molality of the solution
Similarly, the freezing-point depression is calculated using
#DeltaT_"f" = i * K_f * b#
Since you're dealing with a single solution here, i.e. the same molality and van't Hoff factor, you can divide the two equations to get
#(DeltaT_"b")/(DeltaT_"f") = (color(red)(cancel(color(black)(i))) * K_b * color(red)(cancel(color(black)(b))))/(color(red)(cancel(color(black)(i))) * K_f * color(red)(cancel(color(black)(b))))#
This means that you have
#DeltaT_"f" = K_"f"/K_"b" * DeltaT_"b"#
Now, water has a normal boiling point of
#DeltaT_"b" = 100.52^@"C" - 100^@"C" = 0.52^@"C"#
Plug in your values to find the freezing-point depression
#DeltaT_"f" = (1.86 color(red)(cancel(color(black)(""^@"C kg mol"^(-1)))))/(0.512color(red)(cancel(color(black)(""^@"C kg mol"^(-1))))) * 0.52^@"C"#
#DeltaT_"f" = 1.89^@"C"#
Water has a normal freezing point of
You thus have
#T_"f sol" = 0^@"C" - 1.89^@"C" = color(darkgreen)(ul(color(black)(-1.89^@"C")))#
SIDE NOTE The difference between this value and the value given to you in (a) is most likely caused by different values used for