I know the water potential= solute potential + Pressure Potential but how do I f

Question

I know the water potential= solute potential + Pressure Potential but how do I find these if the Sucrose Solution is given. i do have the raw data of initial and final weighy of the given sweet potato. Help please 5. Calculate the water potential and solute (osmotic) potential for each solution analyzed: Solute Potential Sucrose Solution (M) Water Potential 0.0M 0.2M 0.4M 0.6M 0.8M 1.0 M Show the complete calculations for the Water Potential of the 1.0 sucrose solution: Show the complete calculations for the Solute Potential of the 1.0M sucrose solution:

Explanation / Answer

Since the pressure and temperature are not specified , here calculations are made with respect to pressure as atmospheric pressure and temperatures as 20 degree celcius. For solutions at atmospheric pressures, water potential = solute potential

Sucrose solution (M)                   Water potential                      Solute Potential

0.0                                                          0                                         0


solute potential = -iCRT
i = ionisation constant , which is 1 for sucrose solution
C= Molar concentration of sucrose ( 0.2, 0.4, 0.6, 0.8, 1.0)
R= pressure constant (0.0831 l bar/ mole K)
T= Temperature in Kelvin ( celcius + 273)
water potential = solute potential at atmospheric pressure


For 1 M sucrose water potential,
Solute potential = -iCRT
                          = - (1)(1 mol/liter)(0.0831litre bar / mole K)(293 K)
                          = -24.34 bars


For 0.8 M sucrose water potential,
Solute potential = -iCRT
                          = - (1)(0.8 mol/liter)(0.0831litre bar / mole K)(293 K)
                          = - 19.47 bars


For 0.6 M sucrose water potential,
Solute potential = -iCRT
                          = - (1)(0.6 mol/liter)(0.0831litre bar / mole K)(293 K)
                          = -14.60 bars


For 0.4 M sucrose water potential,
Solute potential = -iCRT
                          = - (1)(0.4 mol/liter)(0.0831litre bar / mole K)(293 K)
                          = -9.739 bars


For 0.2 M sucrose water potential,
Solute potential = -iCRT
                          = - (1)(0.2 mol/liter)(0.0831litre bar / mole K)(293 K)
                          = -4.86 bars



To find the isotonic point, in a graph between the percentage change in mass of potato for each sucrose solution , the point that crosses the x axis ( 0 line) will give the isotonic point.
                                                                                                  

                                                                                                      

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