Asked by kay
Which Aqueous solution has the highest normal boiling point?
a.) 0.1 m NaCl
b.) 0.1 m C2H5OH
c.) 0.1 m CaCl2
d.) They all have the same boiling point.
I think that the answer is D, the all have the same boiling point. Is this correct? Because Delta T= K(b)msolute .... and if kb=0.51,
the answer for all of them would be = (0.51)(.1m).
Please explain. I am having trouble with this part of CHEM 2. Thanks!
a.) 0.1 m NaCl
b.) 0.1 m C2H5OH
c.) 0.1 m CaCl2
d.) They all have the same boiling point.
I think that the answer is D, the all have the same boiling point. Is this correct? Because Delta T= K(b)msolute .... and if kb=0.51,
the answer for all of them would be = (0.51)(.1m).
Please explain. I am having trouble with this part of CHEM 2. Thanks!
Answers
Answered by
bobpursley
c) has the highest bp. the bp is raised according to the number of solute particles in solution. NaCl yeilds Na + Cl- (2 particles), alcohol does not dissociate (1 particle), and calcium chloride yields Ca++, and 2 Cl- (three particles).
for CaCl2, bp increase is .52*3*.1
for CaCl2, bp increase is .52*3*.1
Answered by
DrRebel
The answer is 'C' choice. In freezing pt depression and boiling point elevation the number of ions delivered into solution must be multiplied by the molality and constant. This is referred to as the van't Hoff factor.
∆T = (Kb)(m)(i)
∆T = Boiling Pt elevation (colligative property)
Kb = 0.51-degC/molal
i = van’t Hoff Factor
NaCl => (Na^+) + (Cl^-) => van’t Hoff factor (i) = 2 => ‘2’ ions delivered into solution
Ethanol => no ionization occurs = van’t Hoff factor (i) = 1
CaCl2 => (Ca^+2) + 2(Cl^-) = van’t Hoff factor (i) = 3
∆T(NaCl) = (Kb)(m)(i) = (0.51C/m)(0.10m)(2) = +0.102C
∆T(EtOH) = (Kb)(m)(i) = (0.51C/m)(0.10m)(1) = +0.51C
∆T(CaCl2) = (Kb)(m)(i) = (0.51C/m)(0.10m)(3) = +0.153C
∆T = (Kb)(m)(i)
∆T = Boiling Pt elevation (colligative property)
Kb = 0.51-degC/molal
i = van’t Hoff Factor
NaCl => (Na^+) + (Cl^-) => van’t Hoff factor (i) = 2 => ‘2’ ions delivered into solution
Ethanol => no ionization occurs = van’t Hoff factor (i) = 1
CaCl2 => (Ca^+2) + 2(Cl^-) = van’t Hoff factor (i) = 3
∆T(NaCl) = (Kb)(m)(i) = (0.51C/m)(0.10m)(2) = +0.102C
∆T(EtOH) = (Kb)(m)(i) = (0.51C/m)(0.10m)(1) = +0.51C
∆T(CaCl2) = (Kb)(m)(i) = (0.51C/m)(0.10m)(3) = +0.153C
Answered by
Arundhati
C,is the and as cacl2 molecule dissociates into ca+ & 2 cl- ion and has can't Hoff factor 3 so cacl2 has highest boiling point
Answered by
INDIAN
VERY BAD EXPLANATION YOU SHOULD RESTART STUDYING FROM CLASS 1
Answered by
INDIAN
SORRY I HAVE POSTED IT "GALTI SE" DON'T SEE THIS AND THANKS TO EVERY ONE WHO HELP ME TO THIS QUESTION
Answered by
SACHIN
AND MY NAME IS NOT INDIAN MY NAME IS SACHIN A STUDENT OF CLASS 12
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