Question

Determine the osmotic pressure of a solution prepared by dissolving $\text{25}\text{mg}$of ${\text{K}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ in $2$ litre of water at $\text{25}{}^{\text{o}}\text{C}$, assuming that it is completely dissociated.

Answer 1

Hint: The pressure required to stop osmosis is known as osmotic pressure. The osmotic pressure depends upon the molarity of the solution at a given temperature. The osmotic pressure is the product of molarity temperature and the gas constant.

Formula used: $\prod \text{ = }\text{iCRT}$

Complete step by step answer:
The pressure required to stop the flow of solvent towards solvent is known as osmotic pressure.
The formula to calculate the osmotic pressure is as follows:
$\prod \text{ = }\text{iCRT}$
Where,
$\prod$is the osmotic pressure.
$\text{i}$ is the Van't Hoff factor.
C is the molarity.
R is the gas constant.
T is the temperature.
Potassium sulphate is an ionic compound which dissociates in water as follows:
${\text{K}}_2\text{S}{\text{O}}_4\stackrel{{\text{H}}_{\text{2}}\text{O}}{\to }2{\text{K}}^{\text{ + }}+{\text{SO}}_4^{2-}$
Potassium sulphate produces three ions so the value of van’t Hoff factor is $3$.
Convert the amount of potassium sulphate from mg to g as follows:
$\text{1000}\text{mg}\text{ = }\text{1}\text{g}$
$\text{25}\text{mg}\text{ = }0.025\text{g}$
Use the mole formula to determine the number of mole of ${\text{K}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$as follows:
$\text{Mole = }{\displaystyle \frac{\text{Mass}}{\text{Molar}\text{mass}}}$
Molar mass of ${\text{K}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ is $174\text{g/mol}$.
Substitute $174\text{g/mol}$ for molar mass and $0.025\text{g}$ for mass.
$\text{Mole = }{\displaystyle \frac{\text{0}\text{.025}\text{g}}{174\text{g/mol}}}$
$\text{Mole = }0.000144\text{mol}$
Use the mole formula to determine the number of mole of ${\text{K}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$as follows:
$\text{Molarity}\text{ = }{\displaystyle \frac{\text{Moles}\text{of}\text{solute}}{\text{Litter}\text{of}\text{solution}}}$

Substitute $0.000144\text{mol}$ for mole of solute and $2$ liter for volume of the solution.
$\text{Molarity}\text{ = }{\displaystyle \frac{0.000144\text{mol}}{\text{2}\text{L}}}$
$\text{Molarity}\text{ = }7.18\times {10}^{-5}\text{ M}$
Add $\text{273}$ to convert the temperature from $\text{25}{}^{\text{o}}\text{C}$ to kelvin.
$\text{25}{}^{\text{o}}\text{C}\text{ + }\text{273 = }\text{298}\text{K}$
Use the osmotic pressure formula to determine the osmotic pressure as follows:
Substitute $3$ for $\text{i}$, $\text{298}\text{K}$for temperature,$0.0821\text{L}\text{.}\text{atm}\text{.mo}{\text{l}}^{-1}{\text{K}}^{-1}$ for gas constant and $7.18\times {10}^{-5}\text{ M}$ for molarity.
$\prod \text{ = }3\times 7.18\times {10}^{-5}\text{ M}\times 0.0821\text{L}\text{.}\text{atm}.\text{mo}{\text{l}}^{-1}{\text{K}}^{-1}\times 298\text{K}$
$\prod \text{ = }5.27\times {10}^{-3}\text{atm}$

Therefore, the osmotic pressure is $5.27\times {10}^{-3}\text{atm}$.

Note:
The Von’t Hoff factor represents the degree of dissociation or number of ions produced by a compound on dissolution. Molarity is defined as the number of molecules of solute dissolved in a volume of solution. The flow of solvent through a semipermeable membrane towards the solution is known as osmosis. The pressure which causes the movement of solvent from solution to pure solvent is called reverse osmosis.