Question

Given,\[5.8g\] of a non-volatile solute was dissolved in \[100g\]of carbon disulphide\[\left( {C{S_2}} \right)\]. The vapour pressure of the solution was found to be \[190mm\] of $Hg$. Calculate molar mass of solute given the vapour pressure of \[\left( {C{S_2}} \right)\] is \[195mm\] of $Hg$.
[molar mass of \[(C{S_2})\; = {\text{ }}76gmo{l^{ - 1}}\]]

Answer 1

Hint: We need to know that the non – volatile solute will not vaporize or evaporate readily. And it always exhibits a high boiling point and low vapour pressure. Salt, sugar, etc are the examples of non – volatile solutes. In the case of volatile solutes, at the boiling point of the solution, it will produce the vapours. And it has higher vapour pressure than the volatile solutes at the same temperature. Alcohol, mercury, ether, etc are the examples of volatile solutes.

Complete answer:
We need to know that the molar mass of the solute can be found by calculating the number of moles of the solute. Let ‘M’ be the molar mass of the solute which is equal to,
\[number\ of\ moles\ solute = \dfrac{{5.8}}{M} = x\]
The given weight of solute is equal to \[5.8g\].
The formula of mole fraction can be written as,
\[\dfrac{{P_A^0 - {P_S}}}{{P_A^0}} = {X_A}\]
Where, \[P_A^0\] is the vapour pressure of the solute and \[{P_S}\] is equal to the vapour pressure of the solution. Given, \[P_A^0\] is \[195mm\] of $Hg$ and \[{P_S}\] is equal to \[190mm\] of $Hg$. Substitute the given values in above equation,
\[\dfrac{{195 - 190}}{{195}} = {X_A} - - - - \left( 1 \right)\]
Number of moles of solvent \[ = \dfrac{{100}}{{76}} = 1.31 - - - \left( 2 \right)\]
By comparing the equation one and two, I will get the value of ‘x’. Hence,
\[\dfrac{{195 - 190}}{{195}} = \dfrac{x}{{x + 1.31}}\]
\[x = 0.0344moles\]
We have, \[number\ of\ moles\ solute = \dfrac{{5.8}}{M} = x\]
Therefore, substitute the value of x in the above equation,
\[\dfrac{{5.8}}{M} = 0.034\]
On simplification we get,
\[\therefore M = 170.6g/mol\]
Molecular weight of solute is equal to, \[M = 170.6g/mol\]

Note:
We have to know that the molecular weight is equal to the sum of atomic mass of every atom which is present in a compound or the molecule. And the molecular weight is another term of molar mass which is used for the molecular compounds, and it will not depend on the size of the sample. The molecular weight is expressed in terms of kg/mol. And the solution contains mainly two components and that is solute and solvent. The component which is present in smaller particles is known as solute.