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  • Liquid which is decreased by 10 mm HG having no problem of point 2 now what will be the mole fraction of the solvent when it is decreased by 20 mm AG   

A liquid shows a decrease in vapour pressure by 10 mm Hg. What will be the mole fraction of the solvent if the vapour pressure decreases by 20 mm Hg instead?

 

Answers (1)

best_answer

Using a formula, Raoult’s Law - Relative Lowering of Vapour Pressure $$
X_{\text{solvent}}= \frac {P_{0}- P} {P_{0}}
$$

Here, 

  • P0 = Vapour pressure of pure solvent
  • P = Vapour pressure of solution
  • Xsolute = Mole fraction of solute
  • Xsolvent = 1 − Xsolute

Case 1- Decrease in vapour pressure by 10 mm Hg. So, the relative lowering =$$
X_{\text{solvent}}= \frac {100- 90}{100}=0.10
$$

Mole fraction of solute = 0.10

So, mole fraction of solvent = 1 − 0.10 = 0.90

Case 2- Decrease in vapour pressure by 20 mm Hg. So, the relative lowering =$$
X_{\text{solvent}}= \frac {100- 80}{100}=0.20
$$

Mole fraction of solute = 0.20

So, mole fraction of solvent = 1 − 0.20 = 0.80

Posted by

Saniya Khatri

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