drag the appropriate lables to their respective targets .05 NaCl .05 C6H12O6 .05

Question

drag the appropriate lables to their respective targets

.05 NaCl
.05 C6H12O6
.05 CO2
.05 NH4Cl
.05 Al(NO3)3

1. highest boiling point
2. solution with deltaTb = .026C
3. solution that is most strongly dependent upon pressure
4. Solution of ionic compund with the highest freezing point
5. largest van't Hoff Factor

I came up with these van't hoff factors

nacl = 2
c6h12o6 = 1
co2 = 3
NH4Cl = 2  -- i do not know if this one is right
Al(NO3)3 = 4

and the freezing points

nacl = 1.86*2*.05= .186
c6h12o6 = 1.86*1*.05 = .093
co2 = 3*1.86*.05=.279
nh4cl = 2*1.86*.05=.186
alno33 = 4*1.86*.05=.372

and the boiling points

nacl = .512*2*.05=.0512
c6=.512*.05*1=.026
co2 = .0768
nh4cl =2*.512*.05=.0512
alno33 = 4*.512*.05=.1024

so with all of this i got

1. Al(NO3)3
2. C6H12O6
3. CO2
4. NaCl

5 is where i am confused because alno33 has the highest van hoff factor but also the highest boiling
i could see NH4Cl being the largest if i got the vant hoff wrong but even then it would have the highest boiling point as well

if someone could help me out it would be much appriciated.

Explanation / Answer

Here we are calculating only the change in the boiling points using the von't hoff factor and the concentrations of the solution . We are not calculating the original boiling points. This is only the change(increase) in the boiling.This indicates the change is high for ammoniun nitrate but not the actual boiling point. What you have done is correct the von't hoff factor for ammonium chloride is 2. I hope this clered the answer.

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