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

Step-by-step 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* 0.10 M* 0.0821 \text{ L atm}mol^(-1)K^(-1)* 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* 0.10 M* 0.0821 \text{ L atm}mol^(-1)K^(-1)* 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* 0.10 M* 0.0821 \text{ L atm}mol^(-1)K^(-1)* 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* 0.10 M* 0.0821 \text{ L atm}mol^(-1)K^(-1)* 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* 0.10 M* 0.0821 \text{ L atm}mol^(-1)K^(-1)* 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* 0.50 M* 0.0821 \text{ L atm}mol^(-1)K^(-1)* 293 K`

`\pi '=36.08atm`

`\pi < \pi '`

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