Solutions A and B are separated by a membrane that is permeable to urea. Solution A is $10 \mathrm{mM}$ urea, and solution B is $5 \mathrm{mM}$ urea. If the concentration of urea in solution A is doubled, the flux of urea across the membrane will

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[II A]. Flux is proportional to the concentration difference across the membrane, $\mathrm{J}=-\mathrm{PA}\left(\mathrm{C}_{\mathrm{A}}-\mathrm{C}_{\mathrm{B}}\right)$. Originally, $\mathrm{C}_{\mathrm{A}}-\mathrm{C}_{\mathrm{B}}=10 \mathrm{mM}-5 \mathrm{mM}=5 \mathrm{mM}$. When the urea concentration was doubled in solution $\mathrm{A}$, the concentration difference became $20 \mathrm{mM}-$ $5 \mathrm{mM}=15 \mathrm{mM}$ or three times the original difference. Therefore, the flux would also triple. Note that the negative sign preceding the equation is ignored if the lower concentration is subtracted from the higher concentration.

Solutions a and B Are Separated by a Membrane That 10mM10 \mathrm{mM}10mM

Solutions a and B Are Separated by a Membrane That $10 \mathrm{mM}$