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!
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
VERY BAD EXPLANATION YOU SHOULD RESTART STUDYING FROM CLASS 1
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
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
SORRY I HAVE POSTED IT "GALTI SE" DON'T SEE THIS AND THANKS TO EVERY ONE WHO HELP ME TO THIS QUESTION
AND MY NAME IS NOT INDIAN MY NAME IS SACHIN A STUDENT OF CLASS 12
To determine which aqueous solution has the highest normal boiling point, we need to consider the concept of boiling point elevation. The boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure. When a solute is dissolved in a liquid, such as water, the boiling point of the solution increases compared to the pure solvent.
The equation you mentioned, ΔT = K(b)msolute, where ΔT represents the change in boiling point, K(b) is the molal boiling point elevation constant, and msolute is the molality of the solute, is correct.
The molal boiling point elevation constant, K(b), is a characteristic property of the solvent and represents how much the boiling point increases when a solute is added to it. In this case, since all the solutions are aqueous, we can assume that the solvent is water. The value of K(b) for water is approximately 0.51 °C/m.
Now, let's calculate the change in boiling point, ΔT, using the equation ΔT = K(b) * msolute:
a.) For the 0.1 m NaCl solution:
ΔT(a) = (0.51 °C/m) * (0.1 m) = 0.051 °C
b.) For the 0.1 m C2H5OH (ethanol) solution:
ΔT(b) = (0.51 °C/m) * (0.1 m) = 0.051 °C
c.) For the 0.1 m CaCl2 (calcium chloride) solution:
ΔT(c) = (0.51 °C/m) * (0.1 m) = 0.051 °C
As you can see, the change in boiling point is the same for all three solutions. Therefore, the correct answer is D: They all have the same boiling point.
It's important to note that boiling point elevation depends on the concentration of the solute, not the type of the solute, as long as it is a non-volatile solute. In this case, NaCl, C2H5OH, and CaCl2 are all non-volatile solutes, meaning they do not easily vaporize and escape as a gas.