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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Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.