Question

The average osmotic pressure of human body is \(7.8\) bar at \(37^\circ C\). What is the concentration of an aqueous solution of \(NaCl\) that could be used in bloodstream?

• A. \(0.15\;mol\,L^{-1}\)
• B. \(0.30\;mol\,L^{-1}\)
• C. \(0.60\;mol\,L^{-1}\)
• D. \(0.45\;mol\,L^{-1}\)

Hint:

Physiological saline used in medical applications is approximately \(0.15\,M\;NaCl\), which corresponds to nearly \(0.9%\) saline solution. Always include van't Hoff factor for ionic compounds while calculating osmotic pressure.

Answer 1

The Correct Option is A

Concept: Osmotic pressure is given by the relation: \[ \pi = iCRT \] where:
• \( \pi \) = osmotic pressure
• \( i \) = van't Hoff factor
• \( C \) = molar concentration
• \( R \) = gas constant
• \( T \) = temperature in Kelvin For electrolytes like \(NaCl\), dissociation must be considered using van't Hoff factor.

Step 1: Writing the given data.
\[ \pi = 7.8\;bar \] \[ T = 37^\circ C = 310\,K \] For \(NaCl\): \[ NaCl \rightarrow Na^+ + Cl^- \] Therefore, \[ i=2 \] Gas constant: \[ R = 0.083\;L\,bar\,mol^{-1}K^{-1} \]

Step 2: Substituting values in osmotic pressure equation.
\[ \pi = iCRT \] \[ 7.8 = 2 \times C \times 0.083 \times 310 \] \[ 7.8 = 51.46\,C \]

Step 3: Calculating concentration.
\[ C = \frac{7.8}{51.46} \] \[ C \approx 0.15\;mol\,L^{-1} \] Therefore, the concentration of \(NaCl\) solution suitable for bloodstream is: \[ 0.15\;mol\,L^{-1} \]