Question

The volume of a colloidal particle, ${{\rm{V}}_{\rm{c}}}$ as compared to volume of solute particle, ${{\rm{V}}_{\rm{s}}}$ in a true solution could be
A. $\dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,{\rm{1}}{{\rm{0}}^{ - {\rm{3}}}}$
B. $\dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,{\rm{1}}{{\rm{0}}^{\rm{3}}}$
C. $\dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,{\rm{10}}$
D. $\dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,{\rm{1}}{{\rm{0}}^{22}}$

Answer 1

Hint: Particles of the colloidal and true solution are considered as a sphere. So by comparing the volume of the sphere we can determine the ratio of the volume of both colloidal and true solution. The radius is the size of the particles of colloid and true solution.

Complete answer:
A colloid is a mixture in which a phase remains dispersed in the suspension. The size of colloidal particles varies from $10$ to $1000\,\buildrel _{\circ} \over {\mathrm{A}}$.
A true solution is a homogeneous mixture. In the true solution, the size of the solute particles varies from $1$ to $10\,\buildrel _{\circ} \over {\mathrm{A}}$.
The volume of the sphere is, $\dfrac{{\rm{4}}}{{\rm{3}}}{\rm{\pi }}{{\rm{r}}^{\rm{3}}}$
where, ${\rm{r}}$ is the radius of the sphere.

Compare the volume of the colloidal and true solution as follows:
\[\Rightarrow \dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,\dfrac{{\dfrac{{\rm{4}}}{{\rm{3}}}{\rm{\pi r}}_{\rm{C}}^3}}{{\dfrac{{\rm{4}}}{{\rm{3}}}{\rm{\pi r}}_{\rm{S}}^3}}\]
\[\Rightarrow \dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,\dfrac{{{\rm{r}}_{\rm{C}}^3}}{{{\rm{r}}_{\rm{S}}^3}}\]
Substitute the lower values of the radius of the particles,
\[\Rightarrow \dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,\dfrac{{{{\left( {10} \right)}^3}}}{{{1^3}}}\]
$\Rightarrow \dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,{\rm{1}}{{\rm{0}}^{\rm{3}}}$

So, the volume of a colloidal particle, ${{\rm{V}}_{\rm{c}}}$ as compared to volume of solute particle, ${{\rm{V}}_{\rm{s}}}$ in a true solution could be $\Rightarrow \dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,{\rm{1}}{{\rm{0}}^{\rm{3}}}$.

Therefore, option (B) $\dfrac{{{{\rm{V}}_{\rm{C}}}}}{{{{\rm{V}}_{\rm{S}}}}}\,{\rm{ = }}\,{\rm{1}}{{\rm{0}}^{\rm{3}}}$ is correct.

Due to the large surface area, the colloidal state interacts more with the body part which increases the chance of more absorption of the medicine and thus the medicines work more effectively in the colloidal state.
Thus, in colloidal state medicines work effectively because due to the large surface areas the medicine is absorbed easily.

Additional Information:
An example of colloidal is milk, in which the fat particle remains suspended in a water-based solution. The fat particles are not completely soluble in a water-based solution and are large.
Due to the large size colloid solution contains the different phases of suspended particle and suspension medium. Due to the large size of the colloidal particle, the colloidal state cannot easily pass through the various membranes of the body thus it provides more time to the interaction of the drug with the body. So, the drugs are more effective in colloidal form.

Note:
The size of the colloidal particle is ten times more than the size of the solute particles of a true solution. The volume of the colloidal solution is a thousand times more than the volume of the true solution. Due to the large size of suspended particles colloidal states have large surface areas. Whereas in the case of true solutions, the size of particles is very small, the particles of true solutions have a very little surface area.