Question

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

Answer 1

Explanation:

The equation for osmotic pressure, which is:

$\pi=icRT$

where,

$\pi$ = osmotic pressure of the solution

i = Van't hoff factor

c = concentration of solute

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

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

$\pi=2\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K$

$\pi=4.81 atm$

Solution B: 0.10M KBr(aq)

i = 2, (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

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

$\pi '=4.81 atm$

$\pi=\pi '=4.81 atm$ (no flow of water will occur)

b). Solution A: 0.10M $Al(NO_3)_3$ , Solution B: 0.20M $NaNO_3$

Solution A: 0.10M  $Al(NO_3)_3$

i = 4,  (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

$\pi=4\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K$

$\pi=9.62 atm$

Solution B: 0.10M $NaNO_3$

i = 2, (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

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

$\pi '=4.81 atm$

$\pi > \pi '$

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

c). Solution A: 0.10M $CaCl_2$ , Solution B: 0.50M $CaCl_2$

Solution A: 0.10M  $CaCl_2$

i = 3,  (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.10 M

$\pi=3\times 0.10 M\times 0.0821 \text{ L atm}mol^{-1}K^{-1}\times 293 K$

$\pi=7.21atm$

Solution B: 0.50M $CaCl_2$

i = 3, (100% dissociation for electrolytes)

T = 20°C= 20 + 273 K = 293 K

c = 0.50 M

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

$\pi '=36.08atm$

$\pi < \pi '$

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