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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.

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).

gladiator answered 1 year ago

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