This problem has been solved!See the answerThe following solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentrations. Rank the solutions according to their respective osmotic pressures in decreasing order assuming the complete dissociation of ionic compounds.Rank from highest to lowest osmotic pressure. To rank items as equivalent, overlap them.1M MgCl2, 1M KCl, 2M CH3CH2CH2OH, 1M C12H22O11

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Tringa0 Tringa0 · Answer 1 · 2020-01-14T09:11:21+01:00

Explanation:

To calculate the concentration of solute, we use 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]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the solution = T

1) For 1 M [tex]MgCl_2[/tex]:

[tex]MgCl_2\rightarrow Mg^{2+}+2Cl^-[/tex]

i = 3

c = 1 M

[tex]\pi _1=3\times 1M \time RT=3RT[/tex]

2) For 1 M [tex]KCl[/tex]:

[tex]KCl\rightarrow K^{+}+Cl^-[/tex]

i = 2

c = M

[tex]\pi _2=2\times M \time RT=2RT[/tex]

3) For 2 M [tex]CH_3CH_2CH_2OH[/tex]:

i = 1 ( organic molecule)

c = 2 M

[tex]\pi _3=1\times 2M \time RT=2RT[/tex]

4) For 1 M [tex]C_{12}H_{22}O_{11}[/tex]:

i = 1 ( organic molecule)

c = 1 M

[tex]\pi _4=1\times 1M \time RT=1RT[/tex]

Osmotic pressures in decreasing order

[tex]\pi _1>\pi _2=\pi _3>\pi _4[/tex]

1 M [tex]MgCl_2[/tex]>1 M [tex]KCl[/tex]=2 M [tex]CH_3CH_2CH_2OH[/tex]>1 M [tex]C_{12}H_{22}O_{11}[/tex]