Chemistry, 11.10.2019 19:20 july00

Choose the aqueous solution that has the highest boiling point. these are all solutions of nonvolatile solutes and you should assume ideal van't hoff factors where applicable. choose the aqueous solution that has the highest boiling point. these are all solutions of nonvolatile solutes and you should assume ideal van't hoff factors where applicable. a. 0.100 m li₂so₄
b. 0.100 m kno₂
c. 0.200 m c₃h₈o₃
d. 0.060 m li₃po₄e. they all have the same boiling point.

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Answer from: taleiayarbough4544

0.100 m $Li_2SO_{4}$

Explanation:

$\Delta T_b=i\times k_b\times m$

$\Delta T_b$ = Elevation in boiling point

i = Van'T Hoff factor

k_f = boiling point constant

m = molality

1. For 0.100 m $Li_2SO_{4}$

$Li_2SO_4\rightarrow 2Li^{+}+SO_4^{2-}$

i= 3 as it is a electrolyte and dissociate to give 3 ions.

Thus concentration of ions = $3\times 0.100=0.300$

2. For 0.100 m $KNO_{2}$

$KNO_2\rightarrow K^{+}+NO_2^{-}$

i= 2 as it is a electrolyte and dissociate to give 2 ions.

Thus concentration of ions = $2\times 0.100=0.200$

3. For 0.200 m C_3H_8O_3

i= 1 as it is a non electrolyte and do not dissociate to give ions.

4. For 0.060 m Li_3PO_4

$Li_3PO_4\rightarrow 3Li^{+}+PO_4^{3-}$

i= 4 as it is a electrolyte and dissociate to give 4 ions.

Thus concentration of ions = $4\times 0.060=0.24$

Thus as concentration of ions is highest for $Li_2SO_{4}$ and the boiling point will be highest.

Answer from: vanessa7676

0.060 m Li₃PO₄ has the highest boiling point

Further explanation

Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.

Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles because electrolytes break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes.

The term is used in the Solution properties

1. molal

that is, the number of moles of solute in 1 kg of solvent

$\large {\boxed {\bold {m = mole. \frac {1000} {mass \: of \: solvent (in \: grams)}}}$

2. Boiling point and freezing point

Solutions from volatile substances have a higher boiling point and lower freezing points than the solvent

ΔTb = Tb solution - Tb solvent

ΔTb = boiling point elevation

$\rm \Delta T_f = T_fsolvent-T_fsolution$

$\large {\boxed {\boxed {\bold {\Delta Tb \: = \: Kb.m}}}$

$\rm \Delta T_f = K_f \times m$

Kb = molal boiling point increase

Kf = molal freezing point constant

m = molal solution

For electrolyte solutions there is a van't Hoff factor = i

i = 1 + (n-1) α

n = number of ions from the electrolyte

α = degree of ionization, strong electrolyte α = 1, for non electrolytes i = 1

so the boiling point formula becomes:

[tex] \ rm \ Delta T_f = K_b \ times m \ times i [/ tex]

All solutions in the problem have the same solvent -> assuming water (The same [tex] \ rm K_b [/ tex]) so that what affects the value of [tex] \ rm \ Delta T_b [/ tex] is the value of i and m

Assuming the degree of electrolyte ionization α = 1, the magnitude i is determined by the number of ions produced by the electrolyte (n)

a. 0.100 m Li₂SO₄

Li₂SO₄ ---> 2Li ++ SO₄²⁻ → 3 ions

m x i = 0.1 x 3 = 0.3

b. 0.100 m KNO₂

KNO₂ ---> K⁺ + NO₂⁻ → 2 ions

m x i = 0.1 x 2 = 0.2

c. 0.200 m C₃H₈O₃

Non-electrolyte solution, i = 1

ΔTb = Kb .m (based only on concentration m)

m x i = 0.2 x 1 = 0.2 or m only = 0.2

d. 0.060 m Li₃PO₄