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leonid [27]
3 years ago
11

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

the solution ________.
decreases,
is unaffected decreases,
increases decreases,
decreases increases,
increases increases,
decreases
Physics
1 answer:
kykrilka [37]3 years ago
8 0

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

\Delta T_f=K_f\times m

where,

\Delta T_f =depression in freezing point =  

K_f = freezing point constant  

m = molality  ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

  1. If molality of the solution in high the depression in freezing point of the solution will be more.
  2. If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

\frac{p_o-p_s}{p_o}=\chi_{solute}

p_o = Vapor pressure of pure solvent

p_s  = Vapor pressure of solution

\chi_{solute} = Mole fraction of solute

p_s\propto \frac{1}{\chi_{solute}}

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

  1. Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
  2. lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.
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Explanation:

The momentum must be preserved before and after the collision:

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