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A semipermeable membrane separates two aqueous solutions at 20°C. For each of the following cases, name the solution into which a net flow of water (if any) will occur. Assume 100% dissociation for electrolytes.
a. Solution A: 0.10M NaCl(aq) Solution B: 0.10M KBr (aq)
b. Solution A: 0.10M Al(NO3)3 solution B: 0.20M NaNO3
c. Solution A: 0.10M CaCl2 Solution B: 0.50M CaCl2

Respuesta :

Explanation:

The equation for osmotic pressure, which is:

[tex]\pi=icRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution

i = Van't hoff factor

c = concentration of solute

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the solution

a). Solution A: 0.10M NaCl(aq) ,  Solution B: 0.10M KBr (aq)

Solution A: 0.10M NaCl(aq)

i = 2,  (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

[tex]\pi=2\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K[/tex]

[tex]\pi=4.81 atm[/tex]

Solution B: 0.10M KBr(aq)

i = 2, (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

[tex]\pi '=2\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K[/tex]

[tex]\pi '=4.81 atm[/tex]

[tex]\pi=\pi '=4.81 atm[/tex] (no flow of water will occur)

b). Solution A: 0.10M [tex]Al(NO_3)_3[/tex] , Solution B: 0.20M [tex]NaNO_3[/tex]

Solution A: 0.10M  [tex]Al(NO_3)_3[/tex]

i = 4,  (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

[tex]\pi=4\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K[/tex]

[tex]\pi=9.62 atm[/tex]

Solution B: 0.10M [tex]NaNO_3[/tex]

i = 2, (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

[tex]\pi '=2\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K[/tex]

[tex]\pi '=4.81 atm[/tex]

[tex]\pi > \pi '[/tex]

Flow of water will occur from solution B to solution A.

c). Solution A: 0.10M [tex]CaCl_2[/tex] , Solution B: 0.50M [tex]CaCl_2[/tex]

Solution A: 0.10M  [tex]CaCl_2[/tex]

i = 3,  (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

[tex]\pi=3\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K[/tex]

[tex]\pi=7.21atm[/tex]

Solution B: 0.50M [tex]CaCl_2[/tex]

i = 3, (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.50 M

[tex]\pi '=3\times 0.50 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K[/tex]

[tex]\pi '=36.08atm[/tex]

[tex]\pi < \pi '[/tex]

Flow of water will occur from solution A to solution B.

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