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

May 31, 2017

$\Pi = \text{0.9292 atm}$.

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:

$\Pi = i c R T$,

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 $\text{atm}$, then $c$ is in $\text{mol/L}$ when...
• ...the universal gas constant $R$ is $\text{0.082057 L"cdot"atm/mol"cdot"K}$.
• $T$ is the temperature in $\text{K}$. 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")

$=$ $\text{0.005697 mols suc}$

So, the molarity is:

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

$=$ $\text{0.03798 M}$

Therefore, the osmotic pressure at ${25}^{\circ} \text{C" = "298.15 K}$ for the NONELECTROLYTE sucrose is:

$\textcolor{b l u e}{\Pi} = \left(1\right) \left(0.03798 \cancel{\text{mol/L")(0.082057 cancel"L"cdot"atm/"cancel"mol"cdotcancel"K")(298.15 cancel"K}}\right)$

$=$ $\textcolor{b l u e}{\text{0.9292 atm}}$,

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