If the freezing point of blood is -0.52 C, what is its osmotic pressure at 25°C? What is the vapor pressure lowering of blood at this temperature?
The correct answer and explanation is :
To solve for both osmotic pressure and vapor pressure lowering of blood, we need to consider colligative properties.
1. Osmotic Pressure Calculation
Osmotic pressure ((\Pi)) is given by:
[
\Pi = iMRT
]
where:
- ( i ) = Van’t Hoff factor (for blood plasma, we approximate it as 1.9 since it contains a mix of ions and molecules),
- ( M ) = molarity of solutes,
- ( R ) = ideal gas constant = 0.0821 L·atm/(mol·K),
- ( T ) = temperature in Kelvin (25°C = 298 K).
The freezing point depression equation helps us find M:
[
\Delta T_f = i K_f M
]
where:
- ( \Delta T_f = 0.52°C ),
- ( K_f ) for water = 1.86°C·kg/mol.
Solving for M:
[
M = \frac{0.52}{(1.9 \times 1.86)} = 0.148 \text{ mol/kg}
]
Approximating molarity ≈ molality for dilute solutions:
[
\Pi = (1.9)(0.148)(0.0821)(298)
]
[
\Pi \approx 7.3 \text{ atm}
]
2. Vapor Pressure Lowering
Vapor pressure lowering is given by Raoult’s Law:
[
\Delta P = X_{\text{solute}} P_{\text{solvent}}
]
where:
- ( X_{\text{solute}} \approx \frac{0.148}{55.5} = 0.00267 ),
- ( P_{\text{solvent}} ) (vapor pressure of pure water at 25°C) = 23.8 mmHg.
[
\Delta P = (0.00267)(23.8) \approx 0.064 \text{ mmHg}
]
Final Answer:
- Osmotic Pressure ≈ 7.3 atm.
- Vapor Pressure Lowering ≈ 0.064 mmHg.
These values explain how blood maintains osmotic balance and minimizes water loss via evaporation.