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3 years ago

How will the vapor pressure of a solution compare to the pure solvent?

Chemistry

1 answer:

n200080 [17]3 years ago

4 0

Answer: vapor pressure of solution will be less than that of the pure solvent

Explanation:

Vapor pressure of a liquid is defined as the pressure exerted by the vapors in equilibrium with the liquid/solution at a particular temperature.

So, when a non-volatile solute is added to a solvent then its molecules align at the surface of liquid. As a result, less number of solvent molecules will escape from the solution. Thus, there will be decrease in vapors and thus the vapor pressure decrease.

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent

$p_s$ = vapor pressure of solution

$w_2$ = mass of solute

$w_1$ = mass of solvent

$M_1$ = molar mass of solvent

$M_2$ = molar mass of solute

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$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

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