Seawater has a salinity of $3.5\%$, meaning that if you boil away a kilogram of seawater, when you're finished you'll have $35g$of solids (mostly localid="1647507373105" $NaCl$) left in the pot. When dissolved, sodium chloride dissociates into separate $N{a}^{+}$and $C{l}^{-}$ions.

(a) Calculate the osmotic pressure difference between seawater and fresh water. Assume for simplicity that all the dissolved salts in seawater are $NaCl$.

(b) If you apply a pressure difference greater than the osmotic pressure to a solution separated from pure solvent by a semipermeable membrane, you get reverse osmosis: a flow of solvent out of the solution. This process can be used to desalinate seawater. Calculate the minimum work required to desalinate one liter of seawater. Discuss some reasons why the actual work required would be greater than the minimum.

We are given the teh salinity of Seawater is $3.5\%$.

We have to find the osmotic pressure difference between seawater and fresh water and the minimum work required to desalinate one liter of seawater.

We know that,

$V=\frac{m}{\rho }$

Here, m is the mass and $\rho$is the density.

Substitute $m=1kg$and $1000kg/m^3$for$\rho$

$$\begin{gathered} V=\frac{1kg}{1000kg/m^3} \\ ={10}^{-3}{m}^3 \end{gathered}$$

State the molecular weight of NaCl.

$$\begin{gathered} A=23g/mol+35.5g/mol \\ =58.5g/mole \end{gathered}$$

Writing the expression for the number of moles of NaCl,

$$\begin{gathered} n_B=\frac{numberofgrams}{formulaweightofNaCl} \\ =\frac{m}{4} \end{gathered}$$

Substitute$m=35g$and $58.5g/mol$for A,we get

$$\begin{gathered} n_B=\frac{35g}{58.5g/mole} \\ =0.5983moles \end{gathered}$$

Now,

$P_2-P_1=\frac{N_{B}kT}{V}$

Substitute nR for NK in the above expression,

$P_1-P_2=\frac{n_{B}RT}{V}$

Substituting the values, we get

$$\begin{gathered} P_2-P_1=\frac{(0.5983mole)(8.31J/mole.K)(300K)}{{10}^{-3}{m}^3} \\ =\frac{1.49\times {10}^3}{{10}^{-3}}N/m^2 \\ =(1.49\times {10}^{6}Pa)\left(\frac{1atm}{1.01\times {10}^{5}Pa}\right) \\ =14.9atm \end{gathered}$$

Therefore, the osmatic pressure difference between seawater and fresh water is $14.9atm$.

$W=(P_2-P_1)V$

Put role="math" localid="1647756016152" $P_2-P_1=1.49\times {10}^{6}Pa$and ${10}^{-3}{m}^3$ for V,we get.

$$\begin{gathered} W=(14.9\times {10}^{5}Pa)({10}^{-3}{m}^{-3}) \\ =1490J \end{gathered}$$

Therefore, the work needed to desalinate one liter of sea water is $1490J$.