Question

Assuming that seawater has a total ion concentration (also known as colligative molarity) of 1.10 M c , calculate how many liters of seawater are needed to produce 45.8 L of fresh water at 20 ∘ C with an applied pressure of 63.0 bar.

Answer 1

Step-by-step explanation:

It is known that relation between osmotic pressure and temperature is as follows.

`\pi = MRT`

where,
`\pi` = osmotic pressure

M = molarity of solution

R = gas constant

T = Temperature in Kelvin

The given data is as follows.

`\pi` = 63.0 bar, T = (20 + 273) K = 293 K

R = 0.0831 L bar/mol K

Therefore, putting the given data into the above formula as follows.

`\pi = MRT`

or, M =
`(\pi)/(RT)`

M =
`(63.0 bar)/(0.0831 L bar/mol K * 293 K)`

= 2.58 mol/L

Now,
`M_(1) = M_(c)` = 1.10 M, M = 2.58 M

`V_(1)` = ?,
`V_(2) = V_(1) - 45.8 L`

As,
`M_(1)V_(1) = M_(2)V_(2)`

`1.10 * V_(1) = 2.58 (V_(1) - 45.8 L)`

`1.10V_(1) = 2.58V_(1) - 118.16`

`-1.48V_(1)` = -118.16

`V_(1)` = 79.83 L

Thus, we can conclude that the required volume of seawater is 79.83 L.