Question

if P and PS are the vapour pressure of solvent and its solution respectively. N1 and N2 are the mole fraction of solvent and solute respectively then​

Answer 1

Solution:

By using Raoult's law

$\sf{\implies \dfrac{P_{s}-P}{P_{s}} = \dfrac{n_{1}}{(n_{1}+n_{2})}}$

Where,

=> Ps = Vapour pressure of solution.

=> P = Vapour pressure of solvent.

=> n1 = Moles of solute.

=> n2 = Moles of solvent.

Now, we put the value in given formula,

$\sf{\implies \dfrac{P_{s}-P}{P_{s}} = \dfrac{n_{1}}{(n_{1}+n_{2})}}$

$\sf{\implies 1-\dfrac{P}{Ps}=\dfrac{n_{1}}{(n_{1}-n_{2})}}$

$\sf{\implies \dfrac{P}{Ps}=1+\dfrac{n_{1}}{(n_{1}+n_{2})}}$

$\sf{\implies \dfrac{P}{Ps}=\dfrac{n_{2}}{(n_{1}+n_{2})}}$

$\large{\boxed{\boxed{\sf{\implies P = Ps \Bigg(\dfrac{n_{2}}{n_{1}+n_{2}}\Bigg)}}}}$

Extra Info:

Raoult's Law:

• Raoult's law states that for a solution of volatiles liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

Answer 2

Explanation:

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