Posted by ....... on Tuesday, February 5, 2013 at 12:41am.
The concentration of NaCl in solution A is 50mL*(50mM)=250 moles of NaCl per mL
The concentration of NaCl in solution B is 100mL*(100mM)=1,000 moles of NaCl per mL
a) Describe and explain which way the water and the Na+ and Cl- ions would flow?
The ions would flow from solution B to solution A to establish an equal concentration gradient since the membrane is permeable to the ions. Since the membrane is also permeable to water, and there is more water on the B side, water would flow from B to A as well.
b) describe and explain what the volumes and concentrations of the two compartments would be at equilibrium?
The ions and the water concentrations would be equal on both sides of the membrane (i.e., the volumes would be the same as well as the concentrations of ions).
c) Describe and explain how the result would differ if the opening were permeable to water only?
Water would flow from the least concentrated side of the membrane (A) to more concentrated side (B) to establish an equal concentration gradient (i.e., equal mole to water ratios). This will result in A having a lesser volume then B to achieve an equal concentration gradient.
I hope this helps.
I have a typo
The concentration of NaCl in solution A is 50mL*(50mM)=250 moles of NaCl TOTAL not per mL
The concentration of NaCl in solution B is 100mL*(100mM)=1,000 moles of NaCl TOTAL not per mL
Also, the concentration of NaCl in solution B is 100mL*(100mM)=10,000 moles of NaCl TOTAL not 1,000 moles
I totally screwed up on the calculations.
The concentration of NaCl in solution A is 50 x 10^-3L*(50mM)=0.250 moles of NaCl
The concentration of NaCl in solution B is 100 x10^-3L*(100mM)=10 moles of NaCl
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