Chemistry, asked by ShikharMehra, 11 months ago

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

Answers

Answered by itzdevilqueena
12

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

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

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{ \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}

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Answered by jamalshaikh1963
0

Explanation:

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