Question

What will be the osmotic pressure of the cell containing .9% is NaCl against Seawater Which 3.5% NaCl by weight

Someone answered this before, but they didn't take into account the disassosciation of the particles of NaCl I think, and the fact that they made a ratio.

The question is asking for the osmotic pressure of the cell containing .9% is NaCl against Seawater Which 3.5% NaCl by weight. I know when you do this against pure water, its simple the osmotic pressure of the .9% NaCL cell, but how would you do this against something that has a 3.5% solution be weight of salt water.

Answer 1

0.9% NaCl means 0.9 g of NaCl in 99.1 g of water.

3.5% NaCl means 3.5 g of NaCl in 96.5 g of water.

Osmotic pressure due 0.9% NaCl,

P₁ = i x C x R x T

P₁ = 2 x ( 0.9 x 1000 / 58.44 x 99.1 ) x 0.08206 x (273.15+25)

P₁ = 7.604 atm

Osmotic pressure due 3.5% NaCl,

P₂ = i x C x R x T

P₂ = 2 x ( 3.5 x 1000 / 58.44 x 96.5 ) x 0.08206 x (273.15+25)

P₂ = 30.368 atm

Now both the pressures are acting in opposite directions, So

Osmotic pressure in cell = 30.368 - 7.604 = 22.764 atm