Calculate solute of a 0.1M NaCl solution at 25 degrees Celsius. If the concentration of NaCl inside the plant cell is 0.15M, which way will the water diffuse if the cell is placed into the 0.1M NaCl solution?
Answer:
The correct answer is -7.429 bars.
Explanation:
The solute potential inside the cell is calculated by the formula = -iCRT
Here, i refers to the number of ions, C refers to the concentration of solute, R refers to the gas constant, and T is the temperature (K)
The concentration within the cell of the plant is 0.15 M for NaCl, i = 2, R = 0.0831 and T = 298 K or 25 degree C.
Solute potential = - (2 0.15 0.0831 * 298) = -7.429 bar
The way at which the water will diffuse if the cell is placed into the 0.1M NaCl solution is: -7.43 bars.
Using this formula
Solute potential (Ψs) = –iCRT
Where:
i =ionization constant=2 because NaCl forms 2 ions
C =Molar concentration= 0.15 mole/liter
R =Pressure constant = 0.0831 liter bar/mole K
T =Temperature in degrees Kelvin = 273 + °C of solution =273 +25°C=298K
Let plug in the formula
Solute potential (Ψs)=- (2)(0.15mole/liter)(0.0831 liters bars/mole-K)(298 K)
Solute potential (Ψs) = - 7.43 bars
Solute potential (Ψs) = 0 + (-7.43 bars)
Solute potential (Ψs) = -7.43 bars
Inconclusion the way at which the water will diffuse if the cell is placed into the 0.1M NaCl solution is: -7.43 bars.
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