The molecular weight of a solute X is greater than that of solute y. Their equal weights are dissolved separately in the equal quantity of the same solvent which solution will show a great relative lowering of vapour pressure. Why?
Answers
Answer:
Because -
Explanation:
According to Raoult's Law, the “RELATIVE LOWERING OF VAPOUR PRESSURE IS EQUAL TO THE MOLE FRACTION OF THE SOLUTE”.
(P°- P)/P°= x²
where (P°-P)/P°= relative lowering of vapour pressure.
Let us consider TWO solutions are prepared by dissolving “W” g of A and “W” g of B, separately in two vessels containing one litre of water.
Now Mole- fraction of A= (W/X)/[(W/X)+ 55.55], ≈(W/X)/55.55
where 55.55 is the number of moles of water in 1- LITRE or 1- kg of water.
Similarly Mole - fraction of B ≈(W/Y)/55.55
Since X> Y, then mole-fraction of the solution with Molecular weight X- will be less than that that with Molecular weight -Y. Therefore the solution containing the solute “Y” with less Molecular weight will show a greater lowering of Vapour Pressure.