0.1 M solution of solute (non-electrolyte) will have a water potential of

-2.3 bars

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Zero

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2.3 bars

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22.4 bars

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Solution

The correct option is A -2.3 bars
Water potential of pure water is zero. Water potential is less than zero as solute is added to water.
Water potential = -iCRT, where
i = ionization constant
C = Molar concentration
R = Pressure constant, that is 0.0831 litre bar/ mol K
T = Temperature in K, that is 273 K.
In the given problem,C = 0.1 M
Hence, water potential = - 1 x 0.1 x 0.0831 x 273 = -2.3 bars.
Thus, the correct answer is option A.

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