Question

State Raoult's law.How is it useful in determining the molecular?


Answer plz​

Answer 1

Answer:

According to Rault's law, for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. When the solute is non-volatile and non-electrolyte, only the solvent moecules are present in vapour phase and contribute to vapour pressure. Let p1 be the vapour pressure of the solvent x1 be its mole fraction p10 be its vapour pressure in the pure state. Then according to raoult's law, P1 ∝ x1 and p1 ∝ a

Answer 2

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DEFINITION OF RAOULT'S  LAW:

Raoult's Law states that " Relative lowering of Vapour Pressure  of solution containing non-volatile solute is equal to mole fraction of the solute. "

If the solution is not dilute the use formula,

$➠ \dfrac{P^{°} - P^{s}}{P^{°}} =\:X_{solute } \longrightarrow\: Equation\:1$

$➠ X_{solute}\:can\: also\: be\: written\: as$

$➠ X_{solute} = \dfrac{n_{solute} }{n_{solute} +n_{solvent} } \longrightarrow\: Equation\:2$

Substitute equation 2 in equation 1

$\dfrac{P^{°}-P^{s}}{P^{°}} = \dfrac{n_{solute}}{n_{solute}+n_{solvent}}$

If the solution is dilute the use,

$➠ \dfrac{P^{°}-P^{s}}{P^{°}} =\dfrac{n_{solute}}{n_{solvent}}$

$➠ \dfrac{P^{°}-P^{s}}{P^{°}} =\dfrac{\dfrac{w}{m} }{\dfrac{W}{M} }$

$➠ \dfrac{P^{°}-P^{s}}{P^{°}} =\dfrac{w\times M}{m\times W}$

$\dfrac{P^{°}-P^{s}}{P^{°}} =\dfrac{w\times M}{m\times W}$

APPLICATIONS OF RAOULT'S LAW:

Calculations become easier using this equations and the results are more accurate specially when the solutions are not very dilute

If they dilute solution they also easy to determine its vapour pressure by using or knowing the gram molcular weight of given solute.

ℌope ℑt ℌelps