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The 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.
1 M BeCl2
1 M KCl
2 M CH3CH2CH2OH
1 M C12H22O11

Respuesta :

Answer: 1 M [tex]BeCl_2[/tex] > 1 M KCl = 2 M [tex]CH_3CH_2CH_2OH[/tex] > 1 M [tex]C_12H_{22}O_{11}[/tex]

Explanation:

[tex]\pi =iCRT[/tex]

[tex]\pi[/tex] = osmotic pressure = ?

i = vant hoff factor

C= concentration in Molarity  

R= solution constant

T= temperature

a) 1 M [tex]BeCl_2[/tex]

i = 3

[tex]BeCl_2\rightarrow Be^{2+}+2Cl^-[/tex]

Thus Concentration of ions = [tex]3\times 1M=3M[/tex]

b) 1 M KCl

i = 2

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

Thus Concentration of ions = [tex]2\times 1M=2M[/tex]

c) 2 M [tex]CH_3CH_2CH_2OH[/tex]

i = 1 ( as it doesnot dissociate)

Thus Concentration of ions = [tex]1\times 2M=2M[/tex]

d) 1 M [tex]C_12H_{22}O_{11}[/tex]

i = 1  ( as it doesnot dissociate)

Thus Concentration of ions = [tex]1\times 1M=1M[/tex]

Thus order from highest to lowest osmotic pressure is:

1 M [tex]BeCl_2[/tex] > 1 M KCl = 2 M [tex]CH_3CH_2CH_2OH[/tex] > 1 M [tex]C_12H_{22}O_{11}[/tex]

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