The average osmotic pressure of human blood is 7.8 atm at 27°C. The concentration of aqueous $NaCl$ solution that can be used in the blood stream. (Given $R=0.082\, L\, atm\, K^{-1}mol^{-1}$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.16 mol/L
To determine the concentration of an aqueous NaCl solution that can be used in the bloodstream based on the average osmotic pressure, we can use the formula for osmotic pressure:
\(\pi = i C R T\)
Where:
\( \pi \) is the osmotic pressure (in atm).
\( i \) is the van 't Hoff factor (for NaCl, \( i = 2 \) since it dissociates into Na⁺ and Cl⁻).
\( C \) is the concentration of the solution (in mol/L).
\( R \) is the ideal gas constant (\( 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)).
\( T \) is the temperature (in Kelvin).
Given:
\( \pi = 7.8 \, \text{atm} \)
\( T = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
\( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
\( i = 2 \) for NaCl
Rearranging the Formula
We can rearrange the formula to find \( C \):
\(C = \frac{\pi}{i R T}\)
Substituting the Values
\(C = \frac{7.8}{2 \times 0.082 \times 300}\)
Calculating the denominator:
\(2 \times 0.082 \times 300 = 49.2\)
Now, calculate \( C \):
\(C = \frac{7.8}{49.2} \approx 0.16 \, \text{mol/L}\)
Conclusion
Thus, the concentration of aqueous NaCl solution that can be used in the bloodstream is approximately: 0.16 mol/L