What is the osmotic pressure for `"1.95 g"` of sucrose dissolved in `"150 mL"` solution at `25^@ "C"`? `"FW"` `=` `"342.2965 g/mol"`

`Pi = "0.9292 atm"`.

You can read more about osmotic pressure (and other colligative properties) here.

The osmotic pressure `Pi`, the pressure needed to stop the solvent flow from low to high concentration across a semi-permeable membrane, is given by:

`Pi = icRT`,

where:

• `i` is the van't Hoff factor, as seen in freezing point depression and boiling point elevation. It can be approximated by the number of particles per formula unit in solution.
• `c` is the concentration in whatever units are appropriate. In this case, if `Pi` is in `"atm"`, then `c` is in `"mol/L"` when...
• ...the universal gas constant `R` is `"0.082057 L"cdot"atm/mol"cdot"K"`.
• `T` is the temperature in `"K"`.

![http://1.bp.blogspot.com/]()

We are indeed given the correct info to calculate the molarity of the sucrose solution.

`1.95 cancel"g suc" xx "1 mol suc"/(342.2965 cancel"g suc")`

`=` `"0.005697 mols suc"`

So, the molarity is:

`c = "0.005697 mols suc"/(150 xx 10^(-3) "L soln")`

`=` `"0.03798 M"`

Therefore, the osmotic pressure at `25^@ "C" = "298.15 K"` for the NONELECTROLYTE sucrose is:

`color(blue)(Pi) = (1)(0.03798 cancel"mol/L")(0.082057 cancel"L"cdot"atm/"cancel"mol"cdotcancel"K")(298.15 cancel"K")`

`=` `color(blue)("0.9292 atm")`,

since nonelectrolytes have `i = 1`, having hardly any further dissociation at all.