Question

Assuming that the NaCl is completely ionized, calculate how much it will lower the solute potential of the soil at 20°C using the solute potential equation: ѰS = –iCRT where i is the ionization constant (2 for NaCl), C is the molar concentration (in mol/L), R is the pressure constant [R = 0.00831 L • MPa/(mol • K)], and T is the temperature in Kelvin (273 + °C). How much will the solute potential of the soil be lowered at 20°C?

Answer 1

Answer:

Ѱ$S=-4.872MPa$

Explanation:

Hello,

For this exercise, it must be clear that the concentration is not given, nonetheless, one can assume it arbitrarily as 1M, thus, the solute potential turns out into:

Ѱ$S=-iCRT=-2*1\frac{mol}{L} *0.00831\frac{MPa*L}{mol*K} *293.15K$

Ѱ$S=-4.872MPa$

Nevertheless, if you have a different concentration, you just must change the 1M concentration by the one you're given and subsequently modify the answer.

Best regards.