State Raoult's law. How is the molecular mass of a solute determined from lowering of vapour pressure measurement ?

Explanation

Raoult.s Law The French Chemist, Francois Marte Raoult gave a quantitative relationship between the partial pressures and the mole fractions of two components. This relationship is known as Raoult.s law, which states that, "At a given temperature, for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
Let two volatile liquids 1 and 2 have mole fractions as `x_1 and x_2`, respectively. If `p_1 and p_2` are vapour pressures of these components in the solution, then according to Raoult.s law
For component 1, `p_1propx_1 and p_1=p_1^@x_1`
Similarly, for component 2, `p_2=p_2^@x_2`
Where `p_1^@` is the vapour pressure of pure component 1 and `p_2^@` is the vapour pressure of pure component 2 at same temperature.
Determination of Molecular Mass of Solute
Let, `x_1` is the mole fraction of solvent, `x_2` is that of solute, `p_1^@` is the vapour pressure pure solvent and p is that of solution
According to Raoult.s law,
`p_1=p_1^@x_1 or p=p_1=p_1^@x_1`
Since `x_1` is always less than 1, so vapour pressure of the solution will always be less than the vapour pressure of pure solvent `(p_1^@)`
So, lowering in vapour pressure of solvent is
`Deltap_1=p_1^@-p_1=p_1^@-p_1^@x_1=p_1^@(1-x_1)`
But, `x_1=1-x_2 or x_2=1-x_1`
So, `Deltap_1=p_1^@x_2`
or `x_2=(Deltap_1)/(p_1^@)=(p_1^@-p_1)/(p_1^@)`
So, relative lowering in vapour pressure of an ideal solution containing the non-volatile solute is equal to the mole fraction of the solute at a given temperature. Let, W., W, are the masses and M, and M, are the molar masses of the solvent and solute, respectively. If ny moles of solute be dissolved in n moles of the solvent, then the mole fraction if the solute = `n_2/(n_2+n_1)`
For very dilute solutions, `n_2ltltltn_1`
`x_2=(W_2//M_2)/(W_1//M_1)`
So, `(p_1^@-p_1)/(p_1^@)=(W_2M_1)/(W_1M_2) or (p_1^@-p_1)/(p_1^@)=n_2/n_1`
Thus, molar mass of solute can be determined by
`M_2=(W_2M_1)/(W_1((p_1^@-p_1)/p_1^@))`