Question

How is the ore of aluminium concentrated?
(A.) Roasting
(B.) Leaching
(C.) Froth floatation
(D.) Using Wilfley table

Answer 1

Hint: We should know that relative lowering in vapour pressure is one of the four colligative properties. This is directly proportional to the mole fraction of the component present in the less amount in solution.

Complete step by step answer:

We should know that Vapour pressure is the pressure exerted by the vapours over the liquid under the equilibrium conditions at a given temperature. After adding the solute in the solution, the vapour pressure of the solution is found to be lower than that of the pure liquid at a given temperature.

Now we will see and understand how this lowering in vapour pressure is determined mathematically.
Let us assume a binary solution in which the mole fraction of the solvent be ${ X }_{ 1 }$ and that of the solute be ${ X }_{ 2 }$, ${ P }_{ 1 }$ be the vapour pressure of the solvent and ${ P }_{ 1 }^{ 0 }$ be the vapour pressure of the solvent in the pure state.
According to Raoult’s Law:
${ P }_{ 1 }$ = ${ X }_{ 1 }$${ P }_{ 1 }^{ 0 }$…….(1)
The decrease in vapour pressure of the solvent (∆P1) is given by:
$\Delta { P }_{ 1 }$=${ P }_{ 1 }^{ 0 }$-${ P }_{ 1 }$
using equation (1)
$\Delta { P }_{ 1 }$=${ P }_{ 1 }^{ 0 }$-${ X }_{ 1 }$${ P }_{ 1 }^{ 0 }$
$\Delta { P }_{ 1 }$=${ P }_{ 1 }^{ 0 }$(1-${ X }_{ 1 }$)
Since we have assumed the solution to be a binary solution, ${ X }_{ 2 }$=1-${ X }_{ 1 }$
$\Delta { P }_{ 1 }$=${ P }_{ 1 }^{ 0 }$${ X }_{ 2 }$
${ X }_{ 2 }$ = $\Delta { P }_{ 1 }$/${ P }_{ 1 }^{ 0 }$
Or,
$\Delta { P }_{ 1 }$/${ P }_{ 1 }^{ 0 }$ = ${ X }_{ 2 }$
Finally, this equation gives the relative lowering in vapour pressure which is equal to the mole fraction of the solute.
Mole fraction of the solute is also known as the ratio of solute molecules to the total number of molecules in the solution.

Therefore, we can conclude that the correct answer to this question is option C.

Note: The lowering in vapour pressure is due to the fact that after the solute was added to the pure liquid (solvent), the liquid surface now had molecules of both, the pure liquid and the solute. Also, this decrease in vapour pressure depends on the amount of non-volatile solute added in the solution irrespective of its nature and hence it is one of the colligative properties.