Question

The molecular weight of $NaCl$ determined by osmotic pressure method will be:
(A) Same as theoretical value
(B) Higher than theoretical value
(C) Lower than theoretical value
(D) negative

Answer 1

Osmosis is the moment of value of water from an area of high-water concentration to the area of low water concentration across a semipermeable membrane. When water moves across the membrane, and that pressure is known as osmotic pressure.

Complete step by step answer:
Colligative properties are the properties of resolutions that depend only upon the molar concentration of the solution and not on the nature of solution.
As we know, osmotic pressure also depends upon the molar concentration of the solution. Hence, it is also a colligative property.
Osmotic pressure can be used to calculate the molecular mass. As given to us in the question, we need to calculate the molar mass of $NaCl$.
We know that $NaCl$ is a strong electrolyte and it is dissociates into $N{a^ + }$ and $C{l^ - }$
$NaCl \to N{a^ + }\,\,C{l^ - }$
The formula used for osmotic pressure calculation is:
$\pi = iMRT$
Where $\pi$ is the osmotic pressure, $i$ is Van’t Hoff factors, $M$ is molar concentration, $R$ is ideal gas constant and $T$ is temperature ($K$).
In the given formula, $i$ i.e. The Van't Hoff factor is described as the number of ions of solute formed upon dissociation.
In $NaCl$, two ions are formed ($N{a^ + }$ and $C{l^ - }$). Therefore, its value for $i$ is $2$
Now if we see
$\pi = 2MRT$ (because $i = 2$)
Due to the increase in Van’t Hoff’s factor value, value of osmotic pressure in the value of $\pi$, the value of $M$ increases as they are directly proportional.
$\pi \,\alpha \,M$
But, $M = \dfrac{{{\text{number of moles of solute}}}}{{{\text{volume in litre}}}}$
Therefore, $M$ is directly proportional to number of moles ($n$)
In turn, $n = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$
So, the number of moles are inversely proportional to the molar mass of the compound. But keeping in view all the above relations, we can say that $\pi$ or osmotic pressure is increased by two times (because $i = 2$), the molar or molecular mass of $NaCl$ would decrease by two times and become half.
Hence the correct option is C, lower than theoretical value.

Note:
There is another way for calculation of Van’t Hoff’s factor:
$i = 1 + \left( {n - 1} \right)\alpha$
$\alpha$ is the degree of dissociation such as $NaCl$, the value of $\alpha = 1$
So, $i = 1 + \left( {n - 1} \right) \times 1$
$i = 1 + n - 1$
$i = n$
$n =$ Number of moles
From the equation, $NaCl \to N{a^ + }\,\,C{l^ - }$ we can see, one mole of $NaCl$ produces two moles of ions. So, $n = 2$
Therefore, $i = n$, $i = 2$