Question

A mixture of water and acetone boils at 1.25 atm will boil at 348.15K. Given the vapor pressure of acetone (1.58 atm) and water(0.312 atm), what is the composition of the solution in terms of mole fraction of each chemical present

Answer 1

A mixture of water and acetone boils at 1.25 atm will boil at 348.15K. Given the vapor pressure of acetone (1.58 atm) and water(0.312 atm). 5.06 is the composition of the solution in terms of mole fraction of each chemical present.

What is Raoult's Law ?

Raoult's Law is a nonvolatile solute lowers the vapour pressure of the solvent.

It is expressed as:

$P_{\text{soln}} = x_{\text{solvent}} P_{\text{solvent}}$

where,

$P_{\text{sol}$ = vapor pressure of the solution

$x_{\text{solvent}}$ = mole fraction of the solvent

$P_{\text{solvent}}$ = vapor pressure of the pure solvent

Now put the values in above expression, we get

$P_{\text{soln}} = x_{\text{solvent}} P_{\text{solvent}}$

$x_{\text{solvent}} = \frac{P_{\text{soln}}}{P_{\text{solvent}}}$

            $= \frac{1.58}{0.312}$

            = 5.06

Thus from the above conclusion we can say that A mixture of water and acetone boils at 1.25 atm will boil at 348.15K. Given the vapor pressure of acetone (1.58 atm) and water(0.312 atm). 5.06 is the composition of the solution in terms of mole fraction of each chemical present.

Learn more about the Raoult's Law here: https://brainly.com/question/10165688

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