Question

Determine the osmotic pressure of a solution prepared by dissolving $2.5 \times {10^{ - 2}}{\text{ g}}$ of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in ${\text{2 L}}$ of water at ${25^ \circ }{\text{C}}$ assuming that it is completely dissociated.
(${\text{R}} = 0.0821{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$, Molar mass of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ $= 174{\text{ g mo}}{{\text{l}}^{ - 1}}$)

Answer 1

The pressure applied to a pure solvent so that it does not pass into the given solution by osmosis is known as the osmotic pressure. ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ dissociates into two potassium cations and one sulphate anion.

Formula Used: ${\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}$
$\pi = i \times \dfrac{n}{V} \times RT$

Complete step by step answer:
Calculate the number of moles of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ using the equation as follows:
${\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}$
Substitute $2.5 \times {10^{ - 2}}{\text{ g}}$ for the mass of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, $174{\text{ g mo}}{{\text{l}}^{ - 1}}$ for the molar mass of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. Thus,
${\text{Number of moles of }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} = \dfrac{{2.5 \times {{10}^{ - 2}}{\text{ g}}}}{{174{\text{ g mo}}{{\text{l}}^{ - 1}}}}$
${\text{Number of moles of }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} = 1.43 \times {10^{ - 4}}{\text{ mol}}$
Thus, the number of moles of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$are $1.43 \times {10^{ - 4}}{\text{ mol}}$.
Calculate the van’t Hoff factor for ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ as follows:
The number of individual ions and ionic solid dissociates is known as the van't Hoff factor.
${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ dissociates into two potassium cations and one sulphate anion. Thus, the van’t Hoff factor for ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is ${\text{3}}$.
Calculate the osmotic pressure of the solution using the equation as follows:
$\pi = i \times \dfrac{n}{V} \times RT$
Where, $\pi$ is the osmotic pressure,
n is the number of moles of solute,
V is the volume of the final solution in litres,
R is the universal gas constant,
T is the temperature in kelvin,
‘i’ is the van’t Hoff factor of the solute.
Substitute $1.43 \times {10^{ - 4}}{\text{ mol}}$ for the number of moles of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, ${\text{2 L}}$ for the volume of the water, $0.0821{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for the universal gas constant, ${25^ \circ }{\text{C}} + 273 = 298{\text{ K}}$ for the temperature, ${\text{3}}$ for the van’t Hoff factor of ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. Thus,
$\pi = 3 \times \dfrac{{1.43 \times {{10}^{ - 4}}{\text{ mol}}}}{{{\text{2 L}}}} \times 0.0821{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \times 298{\text{ K}}$
$\pi = 5.24 \times {10^{ - 3}}{\text{ atm}}$
Thus, the osmotic pressure of the solution is $5.24 \times {10^{ - 3}}{\text{ atm}}$.

Note: Do not use the temperature value in $^ \circ {\text{C}}$. Convert the temperature from$^ \circ {\text{C}}$ to ${\text{K}}$ using the relation that ${0^ \circ }{\text{C}} = 274{\text{ K}}$. Calculate the number of moles of the solute using the relation that the number of moles is the ratio of mass to molar mass.