Question

derive the relationship between relative lowering of vapour pressure and molare mass of salute ?

Answer 1

$\huge\boxed{ \underline{ \underline{ \bf{Answer}}}}$

 Δ $\frac{P}{P^{0}1 }\:=\frac{n2^{} }{n1^{} }= \:\frac{W2/M2}{W1/M1} = \:\frac{W2M2}{W1M2}$

______________________________________

${ \underline{ \underline { \bf{Explanation}}}}$

W2 g of solute of mass M2 be disolved in W1 g of solvent of molar mass M1

∵ hence

we can write a solution of number of moles of solute  N1 &  &number of moles of solute N2 are given us ,

$=>$   $^{n}1 \:=\frac{W1}{M1} \: and \:^{n} 2\:=\frac{W2}{M2}$

The mole fraction of solute,

$x_{2} \:=\frac{^{n}2}{^{n}1\:+\:^{n} 2}$

    = $\frac{W_{2}/M_{2}}{W_{1}/M_{1}+\:W_{2}/M_{2}}$

    = $\frac{P^{0}_{1}\:P}{P^{0}_{1}}$

where , $P^{0}_{1}$ and p are the vapour pressures of solvent and solution respectively.

∴ Δ$\frac{P}{P^{0}1}\:}\:=\:\frac{P^{0}1\:-\:P}{P^{0}_{1} }$

          = $x_{2}$

         = $\frac{W_{2}/M_{2}}{W_{1}/M_{1}+\:W_{2}/M_{2}}$

For a dilute solutions ,

$^{n} 1$ >> $^{n} 2$

∵ hence,

$^{n} 2$ may be negleted in comparison with $^{n} 1$

                                       

∴ Δ $\frac{P}{P^{0}1 }\:=\frac{n2^{} }{n1^{} }= \:\frac{W2/M2}{W1/M1} = \:\frac{W2M2}{W1M2}$

________________________________________________

Tip :- if you have any queries related to the answer ; mention in the comment box...!

have a good day :)

                               

Answer 2

Explanation:

answer my question of Marathi that is activity Shabd ya vishayavar Suvichar mirdvoon liha

Check in m my profile

plz