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What is relative lowering of vapor pressure? How is it useful in determining the molar mass of a solute?

Answer
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Hint: Raoult’s law implies that the vapor pressure of solvent in a solution at a given temperature is directly proportional to its mole fraction. Relative lowering of vapor pressure is a colligative property. Properties depending upon the mole fractions of solute are known as colligative properties.

Complete step by step answer:
When non-volatile solutes are added in a solution, they reduce the tendency of solvent molecules to escape into the vapor phase. In other words, they reduce the vapor pressure of the solvent molecules. The relative lowering of vapor pressure is the ratio of lowering of vapor pressure to the vapor pressure of the pure solvent.

Let the total vapor pressure of the solution be ‘P’ such that
     P=P1+P2
Where, P1 and P2 are the vapour pressures of the solvent and the solute, respectively.
Since, the solute is non-volatile, its contribution to the total vapour pressure will be zero, i.e. P2=0.

According to Raoult’s law, we can show
     P1x1P1=x1P1o
P1o is the vapour pressure of the pure solvent and x1 is the mole fraction of the solvent in the solution.

If x2 is the mole fraction of the solute in the solution then, we have, x1+x2=1.
Substituting the value of x2 in equation P1=x1P1o and rearranging, we get,P1ox2=P1oP1.
Here, P1oP1 represents lowering of vapor pressure. We finally derive the relative lowering of vapor pressure as:
     x2=P1oP1P1o

Therefore, the relative lowering of vapor pressure of a solution having non-volatile solutes is equal to the mole fraction of the solute.
Determination of molar mass of solute by relative lowering of vapor pressure:
If n2 is the number of moles of solute present and n1 is the number of moles of solvent in the solution then, we can write the above equation as:
     x2=P1oP1P1ox2=n2n1+n2P1oP1P1o=n2n1+n2

We generally assume that the solution is very dilute, i.e. n1+n2n1.
Therefore,
     P1oP1P1o=n2n1

We know that the number of moles is given as, n=given weight (w)molar mass (M)
Therefore, we can write
n2n1=P1oP1P1o=w2M2×M1w1

Thus, the molar mass of the solute, M2=P1oP1P1o×w2w1×M1.

Additional Information:
Colligative property varies inversely with the molar mass of a solute.

Note: Use proper symbols for different physical quantities. Close caution is needed while deducing the expressions. Do not confuse symbols for vapor pressure of the solution (P), vapor pressure of the pure solvent (P1o) and vapor pressure of the solvent in solution (P1).
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