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Sagot :
Answer:
Explanation:
To calculate the solute potential (\( \Psi_s \)) of a 2.0 M sucrose solution at 20 degrees Celsius under standard atmospheric conditions, we need to use the following formula:
\[ \Psi_s = -i \cdot n \cdot R \cdot T \]
Where:
- \( i \) is the ionization constant (for sucrose, which does not ionize in solution, \( i = 1 \)),
- \( n \) is the number of particles into which the solute dissociates (for sucrose, \( n = 1 \)),
- \( R \) is the gas constant (\( 0.0831 \) liter bar per mole per Kelvin),
- \( T \) is the temperature in Kelvin.
Given:
- Concentration of sucrose solution, \( C = 2.0 \) M,
- Temperature, \( T = 20 \) degrees Celsius.
First, convert temperature to Kelvin:
\[ T = 20 + 273.15 = 293.15 \text{ K} \]
Now, calculate the solute potential:
\[ \Psi_s = -1 \cdot 1 \cdot 0.0831 \cdot 293.15 \]
\[ \Psi_s = -24.426 \]
Therefore, the solute potential (\( \Psi_s \)) of a 2.0 M sucrose solution at 20 degrees Celsius under standard atmospheric conditions is approximately \( \boxed{-24.426} \) bars.
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