Question

Dry air was passed successively through a solution of $5\,g$ of a solute in $180\,g$ of water and then through pure water. The loss in weight of solution was $2.5\,g$ and that of pure solvent $0.04\,g$. The molecular weight of the solute is:
A. $31.25$
B. $3.125$
C. $312.5$
D. None of the above

Answer 1

Hint: Molecular mass is another name for molecular weight and the total mass of the atoms that are make a molecules that is known as its molecular mass. We must use the Ostwald-Walker hypothesis to ascertain the solute's molecular weight.

Formula Used:According to the theory of Ostwald-Walker the formula is written as:
$\dfrac{{\Delta P}}{P} = \dfrac{{{w_2}}}{{{w_1} + {w_2}}}$
Here, ${\Delta P}$ is change in pressure, P is total pressure, ${w_2}$ is the loss in mass of solvent and ${w_1}$ is the loss in mass of solution .

Complete step-by-step answer:In the question, we have given a solution of $5\,g$ of a solute in $180\,g$ of water present in the dry air,
And, the loss in weight of the solution is ${w_1} = 2.5\,g$ and a pure solvent is ${w_2} = 0.04\,g$,
According to the theory of Ostwald-Walker the formula is written as:
$\dfrac{{\Delta P}}{P} = \dfrac{{{w_2}}}{{{w_1} + {w_2}}}$
Now, substitute the given information in the above formula, then we have:
$\dfrac{{\Delta P}}{P} = \dfrac{{0.04}}{{2.5 + 0.04}} = 0.0157\,\,\,....(i)$

Raoult's law defines a phenomenological relation that presumes the ideal behaviour and based on the straightforward microscopic assumption that the intermolecular forces that are between the dissimilar molecules is equal to that for identical molecules.
These conditions are considered the best case scenario. This is connected to the ideal gas law, a limiting law that applies when the molecular forces approach zero. In light of Raoult's law, we now have:
$\dfrac{{\Delta P}}{P} = \dfrac{{\dfrac{Mass\,of\,solute}{Molar\,Mass\,of\,solute}}}{ \dfrac{{Mass\,of\,Solvent}}{{Molar\,Mass\,of\,Solvent}}}$
$\dfrac{{\Delta P}}{P} = \dfrac{{\dfrac{5}{M}}}{{ \dfrac{{180}}{{28}}}}\,\,\,....(ii)$
Compare the equation $(i)$from $(ii)$, then we obtain:
$0.0157 = \dfrac{{\dfrac{5}{M}}}{{\dfrac{{180}}{{28}}}} \\$
 $ \Rightarrow \left( {\dfrac{5}{M}} \right) = 10 \times 0.0157 \\$
 $\Rightarrow M = \dfrac{{5}}{{10 \times 0.0157}} \\$
 $\Rightarrow M ≈ 31.25\,g \\$
Therefore, the molecular weight of the solute is $31.25\,g$.

Option ‘A’ is correct

Note:The method we used in this solution is used for calculating the molar mass of non-volatile solute. The equation given by Ostwald-Walker and by Raoult's law for lowering of vapour pressure is used to calculate the molar mass of solute when solvent is known.