Question

In: Biology

A membrane separates two solutions. Solution C is 300 mL of 0.2 M NaCl, and Solution...

A membrane separates two solutions. Solution C is 300 mL of 0.2 M NaCl, and Solution D is 500 mL of 0.6 M NaCl. The membrane is permeable to water, sodium ions, and chloride ions. After 24 hours, what will be the chloride ion concentration in Solution C? Show your calculations.

Solutions

Expert Solution

Solution C is 300 mL of 0.2 M NaCl

Molecular weight of NaCl is 58.44 g/mol.

1 M NaCl has 58.44 g/1000 ml

0.2 M NaCl has 11.688 g/1000 ml (58.44 x 0.2)

0.2 M NaCl has 3.5064 g/300 ml (11.688 x 300/1000 = 3.5064)

So, solution C has 3.5064 g of NaCl.

Solution D is 500 mL of 0.6 M NaCl.

1 M NaCl has 58.44 g/1000 ml

0.6 M NaCl has 35.064 g/1000 ml (58.44 x 0.6)

0.2 M NaCl has 17.532 g/500 ml (35.064 x 500/1000 = 17.532).

So, solution D has 17.532 g of NaCl.

After 24 hours, both solution C and D will have equal volume of water and equal concentration of NaCl, as the membrane is permeable to water, Na+, and Cl- ions.

Total volume is 800 ml (300+500). So, each solution would have 400 ml.

Total quantity of NaCl is 21.0384 g (3.5064 + 17.532) in 800 ml. This means 0.45 M concentration.

800 ml has 21.0384 g NaCl (i.e. 26.298 g/L).

If 1 M NaCl has 58.44 g/1000 ml, then the concentration of the solution at 26.298 g/L is 0.45 M (26.298/58.44 = 0.45).


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