Question

Identify the right option:
The average osmotic pressure of human blood is $7.8\;bar$ at $3{7^0}\;C$. What is the concentration of an aqueous NaCl solution that could be used in the bloodstream?
A) $0.15\;mol/L$
B) $0.30\;mol/L$
C) $0.45\;mol/L$
D) $0.60\;mol/L$

Answer 1

Hint: We can define osmotic pressure as a force applied by a solution which passes through a semipermeable membrane during the process of osmosis. This force resists the solution to pass back through the semipermeable membrane.
We can solve this problem using this formula, $\prod = CRT\; \times \;i$
Where, \[\prod = \] Osmotic pressure
\[C{\text{ }} = \] Concentration of \[NaCl\]
\[R{\text{ }} = \] Universal gas constant
\[T{\text{ }} = \] Temperature
\[i{\text{ }} = \] Van’t Hoff factor

Complete step by step answer:
In simple words, we can define osmotic pressure as the minimum pressure applied by a solution to resist the inward flow of the pure solvent through the semipermeable membrane.
We know the osmotic pressure of blood is because of the presence of serum albumin, protein etc found in the blood plasma. These help to maintain osmotic pressure in the vessels and tissues.
Now, as we know the osmotic pressure of human blood is \[7.8{\text{ bar}}\] at \[{\text{3}}{{\text{7}}^0}{\text{C}}\].
So, using the formula, $\prod = CRT\; \times \;i$, we can easily get the concentration of NaCl used in the bloodstream.
We know that,\[{\text{NaCl}}\] dissociates in the aqueous medium in this way,

\[{\text{NaCl}} \to {\text{N}}{{\text{a}}^{\text{ + }}}\;{\text{ + }}\;{\text{C}}{{\text{l}}^{\text{ - }}}\]

Van’t Hoff factor \[\left( i \right){\text{ }} = 2\]

\[{\text{R = 0}}{\text{.0821 L}}{\text{.atm}}{\text{.}}{{\text{K}}^{ - 1}}.{\text{mo}}{{\text{l}}^{ - 1}}\]
${\Pi _{blood}}$\[ = {\text{ }}7.8{\text{ }}bar\]
\[{\text{T = 3}}{{\text{7}}^0}{\text{C = 310 K}}\]
We know,
${\Pi _{blood}} = {C_{{\text{NaCl}}}}R{\text{ $\times$ T $\times$ i}}$
$7.8{\text{ bar}} = {C_{Na{\text{C}}l}}{\times}{\text{0}}{\text{.0821 L}}{\text{.atm}}{\text{.}}{{\text{K}}^{ - 1}}.{\text{mo}}{{\text{l}}^{ - 1}}{\times} 310{\text{ K}} {\times}2$

\[{C_{Na{\text{C}}l}}{\text{ = }}\dfrac{{7.8\;{\text{bar}}}}{{{\text{0}}{\text{.0821 L}}{\text{.atm}}{\text{.}}{{\text{K}}^{ - 1}}.{\text{mo}}{{\text{l}}^{ - 1}}{\text{ $\times$ 310K $\times$2}}}}\]
${{\text{C}}_{{\text{NaCl}}}}{\text{ = 0}}{\text{.153mol/L}}$
The concentration of an aqueous \[NaCl\] solution that is used in blood stream is $0.15\;mol/L$

Hence, the option \[\left( {\mathbf{A}} \right)\] is Correct.

Note: Osmotic pressure exerted by blood stream is basically colloidal osmotic pressure.
It is colloidal because the protein, serum albumin replaces water molecules in the blood plasma which creates a deficit of water in blood plasma. It creates a water deficit in the tissues.