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

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:
${{\text{K}}_2}{\text{S}}{{\text{O}}_4}\mathop \to \limits^{{{\text{H}}_{\text{2}}}{\text{O}}} \,2\,{{\text{K}}^{\text{ + }}}\, + \,{\text{SO}}_4^{2 - }$
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:
${\text{Mole = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
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.
${\text{Mole = }}\,\dfrac{{{\text{0}}{\text{.025}}\,{\text{g}}}}{{174\,{\text{g/mol}}}}$
${\text{Mole = }}\,0.000144\,{\text{mol}}$
Use the mole formula to determine the number of mole of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$as follows:
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Litter}}\,{\text{of}}\,{\text{solution}}}}$

Substitute $\,0.000144\,{\text{mol}}$ for mole of solute and $2$ liter for volume of the solution.
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{\,0.000144\,{\text{mol}}}}{{{\text{2}}\,{\text{L}}}}$
${\text{Molarity}}\,{\text{ = }}\,7.18 \times {10^{ - 5}}{\text{ M}}$
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.