Question

What is the osmotic pressure of an aqueous solution of 1.64 g of \({\bf{Ca}}\left( {{\bf{NO_3}}} \right){\bf{_2}}\) in water at \({25^{\bf{o}}}{\bf{C}}\)? The volume of the solution is 275 ml.

1. Outline the steps necessary to answer the question.
2. Answer the question

Short answer

The Osmotic pressure of an aqueous solution of 1.64g of \({\bf{Ca}}\left( {{\bf{NO_3}}} \right){\bf{_2}}\) in water at \({25^{\bf{o}}}{\bf{C}}\) is 0.89atm.

Step-by-step solution

Osmotic Pressure  

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.

Subpart (a)

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

π = 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

Subpart (b)

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

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