91Ó°ÊÓ

The pressure that might must be applied to a pure solvent to forestall it from passing into a given solution by osmosis, is usually wont to express the concentration of the answer.

Mass of \({\bf{Ca}}\left( {{\bf{NO_3}}} \right){\bf{_2}}\) = 1.64g

Molar Mass of\({\bf{Ca}}\left( {{\bf{NO_3}}} \right){\bf{_2}}\)= 164g/mole

\({\bf{Number}}\,{\bf{of}}\,{\bf{moles = }}\frac{{{\bf{Mass}}\,{\bf{of}}\,{\bf{solute}}}}{{{\bf{Molar}}\,{\bf{Mass}}}}\)

Temperature = \({25^{\bf{o}}}{\bf{C}}\) = 298K

Volume of Solution= 275ml= 0.275L

This is important because the equation we use to determine Osmotic Pressure is

\({\bf{\Pi = iMRT}}\), where

Ï€ = Osmotic pressure

i = van't Hoff's factor

M = Molarity in mole/ L

\(R{\rm{ }} - {\rm{ }}Perfect{\rm{ }}gas{\rm{ }}constant,{\rm{ }}at{\rm{ }}0.08206{\rm{ }}L\;atm/mol\;K\)

T- Temperature in K

Mass of \({\bf{Ca}}\left( {{\bf{NO_3}}} \right){\bf{_2}}\) = 1.64g

Molar Mass of \({\bf{Ca}}\left( {{\bf{NO_3}}} \right){\bf{_2}}\)= 164g/mol

\(\begin{align}Number\,of\,moles &= \frac{{Mass\,of\,solute}}{{Molar\,Mass}}\\Number\,of\,Moles\, &= \,\frac{{1.64\,g}}{{164\,g/mol}}\\\, &= \,0.01\,mol\end{align}\)

\(\begin{align}Temperature{\rm{ }} &= \;25^\circ C\\ &= \left( {273 + 25} \right)K\\ &= {\rm{ }}298K\end{align}\)

\(\begin{align}Volume{\rm{ }}of{\rm{ }}Solution &= {\rm{ }}275ml\\ &= {\rm{ }}0.275L\end{align}\)

Osmotic Pressure is

\(\begin{align}\Pi {\bf{ }} &= {\bf{ }}iMRT\\ &= i \times n/V \times R \times T\\\Pi &= 1 \times 0.01/0.275 \times 0.0821 \times 298\\\Pi &= 0.89atm\end{align}\)