Question

a) Predict the osmotic pressure order for the following( assume salts are $100$ dissociated).
I.$0.1M$ Urea
II.$0.1M$ $NaCl$
III.$0.1M$ $N{{a}_{2}}S{{O}_{4}}$
IV.$0.1M$ $N{{a}_{3}}P{{O}_{4}}$
b) If equal volume of all these solutions are mixed then calculate the osmotic pressure of the net resultant solution obtained at $300K$ .

Answer 1

Osmotic pressure is defined as the minimum amount of pressure that has to be applied to a solution so that to halt the flow of solvent molecules through a semipermeable membrane. It is said to be a colligative property and depends upon the concentration of solute particles in the solution.

Complete step-by-step answer: a)To calculate osmotic pressure, the formula used is as follows:
$\pi =iCRT$
Where, $\pi$ is denoted as the osmotic pressure
$i$ is denoted as van't hoff factor
$C$ is the molar concentration of the solute in the solution
$R$ is denoted as the universal gas constant
$T$ is the temperature
Van't hoff factor $\left( i \right)$ is defined as the number of particles dissociated.
Let us calculate the osmotic pressure order for the following molecules:
I.$0.1M$ urea
It is given that the molar concentration $\left( C \right)$ is $0.1M$
As urea does not dissociate and associate. Therefore, its van't hoff factor $\left( i \right)$ is $1$
Now applying the formula
$\pi =iCRT$
Substituting the value in the above formula we get,
$\pi =1\times 0.1\times RT$
$\pi =0.1RT$
II. $0.1M$ $NaCl$
It is given that the molar concentration $\left( C \right)$ is $0.1M$
Dissociation of $NaCl$-
$NaCl\to N{{a}^{+}}+C{{l}^{-}}$
Hence, the van't hoff factor $\left( i \right)$ is $2$
Now applying the formula
$\pi =iCRT$
Substituting the value in the above formula we get,
$\pi =2\times 0.1\times RT$
$\pi =0.2RT$
III. $0.1M$ $N{{a}_{2}}S{{O}_{4}}$
It is given that the molar concentration $\left( C \right)$ is $0.1M$
Dissociation of $N{{a}_{2}}S{{O}_{4}}$-
$N{{a}_{2}}S{{O}_{4}}\to 2N{{a}^{+}}+SO_{4}^{-}$
Hence, the van't hoff factor $\left( i \right)$ is $3$
Now applying the formula
$\pi =iCRT$
Substituting the value in the above
formula we get,
$\pi =3\times 0.1\times RT$
$\pi =0.3RT$
IV. $0.1M$ $N{{a}_{3}}P{{O}_{4}}$
It is given that the molar concentration $\left( C \right)$ is $0.1M$
Dissociation of $N{{a}_{3}}P{{O}_{4}}$-
$N{{a}_{3}}P{{O}_{4}}\to 3N{{a}^{+}}+PO_{4}^{-}$
Hence, the van't hoff factor $\left( i \right)$ is $4$
Now applying the formula
$\pi =iCRT$
Substituting the value in the above formula we get,
$\pi =4\times 0.1\times RT$
$\pi =0.4RT$
The order of osmotic pressure is
$IV>III>II>I$

b) To calculate the osmotic pressure of the net resultant solution, firstly we need to calculate the $MV$ for each solute.
$MV$
Where, $M$ is the molar concentration
$V$ is the volume
Now as we know that the volume of every solvent is equal.
Hence,
Urea:
$\Rightarrow {{M}_{1}}{{V}_{1}}$
The molar concentration of urea is $0.1$
$\Rightarrow 0.1V$
-$NaCl$ :
$\Rightarrow {{M}_{2}}{{V}_{2}}$
The molar concentration of $NaCl$is $0.2$
$\Rightarrow 0.2V$
-$N{{a}_{2}}S{{O}_{4}}$:
$\Rightarrow {{M}_{3}}{{V}_{3}}$
The molar concentration of $N{{a}_{2}}S{{O}_{4}}$is $0.3$
$\Rightarrow 0.3V$
-$N{{a}_{3}}P{{O}_{4}}$
$\Rightarrow {{M}_{4}}{{V}_{4}}$
The molar concentration of $N{{a}_{3}}P{{O}_{4}}$is $0.4$
$\Rightarrow 0.4V$
Now, we will add these values we get,
${{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}+{{M}_{3}}{{V}_{3}}+{{M}_{4}}{{V}_{4}}={{M}_{5}}{{V}_{5}}$
Now substituting the values we get,
$\Rightarrow 0.1V+0.2V+0.3V+0.4V={{M}_{5}}\times 5V$
$\Rightarrow 1.0V=M\times 5V$
$\Rightarrow {{M}_{5}}=0.2$
As we know to calculate the osmotic pressure, the formula is-
$\pi =iCRT$
Where, $\pi$ is denoted as the osmotic pressure
$i$ is denoted as van't hoff factor
$C$ is the molar concentration of the solute in the solution
$R$ is denoted as the universal gas constant
$T$ is the temperature
It is given that the molar concentration (M) is $0.5$ , temperature(T) is $300K$ and universal gas constant (R) is $0.0821$
Now substituting this value we get,
$\pi =0.2\times 0.0821\times 300$
$\pi =4.926$
Hence, the osmotic pressure is $4.926$

Note: Van't hoff factor is defined as the ratio of the concentration of the particles formed when we dissolve a substance to the concentration of the substance by its mass. The value of van't hoff factor equals to one for non electrolytic substance dissolved in water.