Question

Calculate the amount of \[CaC{l_2}\](van’t Hoff factor \[i = 2.47\]) dissolved in \[2.5\;L\] of water so that its osmotic pressure at \[300K\] is \[0.75\;atmosphere\]. Given: Molar mass of \[CaC{l_2}\]is\[111{\text{ }}g.mo{l^{ - 1}}\], \[R{\text{ }} = {\text{ }}0.082{\text{ }}L.atm.{K^{ - 1}}mo{l^{ - 1}}\]

Answer 1

Hint: Osmotic pressure refers to the force exerted by a solution which passes through a semipermeable membrane by means of osmosis (force required to resist the solution from passing back through the surface. An example of osmotic pressure is the process to filter water.

Step by step answer: Osmotic pressure of a solution is directly related to its concentration while inversely proportional to its volume at a constant temperature. Thus, Van't Hoff's Charle's law states that the osmotic pressure of a dilute solution is directly related to its absolute temperature.

The osmotic pressure can be calculated from the Van't Hoff's equation stated below:
$\pi = iMRT$

Here, i stands for Van't Hoff's factor, M refers to the molar concentration, R stands for the ideal gas constant, T is the temperature and π refers to the osmotic pressure.
In the question, we are given the following information:

\[\begin{array}{*{20}{l}}
  {\pi {\text{ }} = \;0.75{\text{ }}atmosphere} \\
  {V{\text{ }} = {\text{ }}2.5L} \\
  {i{\text{ }} = {\text{ }}2.47} \\
  {T{\text{ }} = {\text{ }}300K} \\
  {R{\text{ }} = {\text{ }}0.0821{\text{ }}L.atm.{k^{ - 1}}.mo{l^{ - 1}}}
\end{array}\]

Molarity of a solution can be calculated from the following formula:
$M = \dfrac{n}{V}$

M = molarity, n = number of moles of solute, V = volume of solution
Let us assume the amount of \[CaC{l_2}\] dissolved in \[2.5\;L\] of water is x g.
The molar mass of solute i.e. \[CaC{l_2}\]is given as \[111{\text{ }}g.mo{l^{ - 1}}\].

‘n’ can be further calculated as: $n = \dfrac{{mass{\text{ }}of{\text{ }}solute}}{{molarmass{\text{ of }}solute}}$$ = \dfrac{x}{{111}}$

Now, substitute the values in Van’t Hoff’s equation:
$
  0.75 = 2.47 \times \dfrac{x}{{111}} \times \dfrac{1}{{2.5}} \times 0.0821 \times 300 \\
  \therefore x = 3.42g \\
$
Hence, the amount of \[CaC{l_2}\] dissolved in \[2.5\;L\] of water is 3.42 g


Note: Osmotic pressure is considered to be the basis of filtration. The water that has to be purified is kept in a chamber and usually placed under the pressure higher than the osmotic pressure that is caused by water as well as the dissolved solutes. Some portion of the chamber opens to a semi-permeable membrane that allows water molecules to pass through, but restrict solute particles. Reverse osmosis is used to desalinate fresh water from salty water.