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At 20ºC, a plant cell containing 0.6 M glucose is in equilibrium with its surrounding solution containing 0.5 M glucose in an open container. What is the cell's ψ P?
    Correct answer is '2.4'. Can you explain this answer?
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    At 20ºC,a plant cell containing 0.6 M glucose is in equilibrium w...
    Surrounding solution: ψ = ψ P + ψ S = 0 bars + (1)(0.5 mol L)(0.0831 L*bars/ mol*K)(293 K) = -12.2 bars. Cell at equilibrium: -12.2 bars = ψ P + (1)(0.6 molL)(0.0831 L *bars mol*K)(293 K) = ψ P + (-14.6 bars) ψ P = -12.2 bars — (-14.6 bars) = 2.4 bars
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    At 20ºC,a plant cell containing 0.6 M glucose is in equilibrium w...
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    At 20ºC,a plant cell containing 0.6 M glucose is in equilibrium with its surrounding solution containing 0.5 M glucose in an open container. What is the cell's ψP?Correct answer is '2.4'. Can you explain this answer?
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