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Therefore, the water potential of the sugar water is -4.0 bars [Ψ = 0 bars +(-4.0) bars]. Since free water always flows towards the solution with a lower water potential, the flow of water would be outside of the cell. 3. The original cell from question # 1 is placed in a beaker of sugar water with Ψ S = -0.15 MPa (megapascals).

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