Question

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.

A. 1 M MgCl2
B. 1 M KCI
C. 1 M C12
D. H22011
1. Highest osmotic pressure
2. Lowest osmotic pressure

Answer 1

Answer: 1. Highest osmotic pressure : 1 M
`MgCl_2`

2. Lowest osmotic pressure: 1 M
`C_(12)H_(22)O_(11)`

Step-by-step explanation:

`\pi=i* C* R* T`

`\pi`= osmotic pressure

i = Van'T Hoff factor

C = concentration

T = Temperature

1. For 1M
`MgCl_2`

, i= 3 as it is a electrolyte and dissociate to give 3 ions.

`MgCl_2\rightarrow Mg^(2+)+2Cl^(-)`

2. For 1 M
`KCl`

, i= 2 as it is a electrolyte and dissociate to give 2 ions.

`KCl\rightarrow K^(+)+Cl^(-)`

3. For 1M
`C_(12)H_(22)O_(11)`

, i= 1 as it is a non electrolyte and do not dissociate.

Thus as vant hoff factor is highest for
`MgCl_2` thus osmotic pressure is highest for 1 M
`MgCl_2`

And as vant hoff factor is lowest for
`C_(12)H_(22)O_(11)` thus osmotic pressure is lowest for 1 M
`C_(12)H_(22)O_(11)`