The vapor pressure of pure water at $75^\circ$is $296torr$ then the vapour pressure of lowering due to $0.1m$ solute is:
A.$0.533torr$
B.$0.296torr$
C.$0.333torr$
D.$0.428torr$
Answer
637.5k+ views
Hint: Raoult’s law states that partial vapour pressure of a solvent in a solution or in mixture is equal or to the vapour pressure of the pure solvent which is multiplied by its mole fraction in the solution. Raoult's law is useful in describing ideal solution
Complete step by step solution:
We know that,
Raoult’s law is given as,
\[{{\text{P}}_{{\text{solvent}}}} = {{\text{X}}_{{\text{solution}}}}{ \times P}_{{\text{Solvent}}}^{\text{0}}\] ……(1)
Here,
\[{{\text{P}}_{{\text{solvent}}}}\]= vapour pressure of solvent
\[{{\text{X}}_{{\text{solution}}}}\]= mole fraction of the solvent
\[{\text{P}}_{{\text{Solvent}}}^{\text{0}}\]= vapour pressure of solvent
It is given to us that,
Vapor pressure of pure water = \[{\text{296}}\,{\text{torr}}\]
Temperature = \[75^\circ {\text{C}}\]
Amount of Solute = \[{\text{0}}{\text{.1}}\,{\text{m}}\]
Number of moles of water \[{\text{ = }}\dfrac{{1000}}{{18}}\]
Using the given values in equation (1), we will now calculate the value of \[{{\text{P}}_{{\text{solvent}}}}\]
\[
{{\text{P}}_{{\text{solvent}}}} = {{\text{X}}_{{\text{solution}}}}{ \times P}_{{\text{Solvent}}}^{\text{0}} \\
{{\text{P}}_{{\text{solvent}}}} = \left( {\dfrac{{0.1}}{{0.1\, + \,1000/18}}} \right) \times 296 = \,0.533\,{\text{torr}} \\
\]
Therefore, the correct option is A.
Additional information:
Few limitations to Raoult’s law are:
1)Raoult's law is useful for describing ideal solutions. However, ideal solutions are rare and hard to find.
2)The negative deviation is seen when the vapour pressure is lower than that of the expected from Raoult's law. A positive deviation is seen when the cohesion between the similar molecules is greater or exceeds the adhesion between dissimilar molecules. Both components of the mixture can easily escape from the solution.
Note: Raoult's law is valid only in ideal solutions. In an ideal solution, the solvent-solute interaction is the same as that of the solvent - solvent or solute - solute interaction. This tells that both the solute and the solvent take the same amount of energy in order to escape to the vapour phase when they exist in their pure states.
Complete step by step solution:
We know that,
Raoult’s law is given as,
\[{{\text{P}}_{{\text{solvent}}}} = {{\text{X}}_{{\text{solution}}}}{ \times P}_{{\text{Solvent}}}^{\text{0}}\] ……(1)
Here,
\[{{\text{P}}_{{\text{solvent}}}}\]= vapour pressure of solvent
\[{{\text{X}}_{{\text{solution}}}}\]= mole fraction of the solvent
\[{\text{P}}_{{\text{Solvent}}}^{\text{0}}\]= vapour pressure of solvent
It is given to us that,
Vapor pressure of pure water = \[{\text{296}}\,{\text{torr}}\]
Temperature = \[75^\circ {\text{C}}\]
Amount of Solute = \[{\text{0}}{\text{.1}}\,{\text{m}}\]
Number of moles of water \[{\text{ = }}\dfrac{{1000}}{{18}}\]
Using the given values in equation (1), we will now calculate the value of \[{{\text{P}}_{{\text{solvent}}}}\]
\[
{{\text{P}}_{{\text{solvent}}}} = {{\text{X}}_{{\text{solution}}}}{ \times P}_{{\text{Solvent}}}^{\text{0}} \\
{{\text{P}}_{{\text{solvent}}}} = \left( {\dfrac{{0.1}}{{0.1\, + \,1000/18}}} \right) \times 296 = \,0.533\,{\text{torr}} \\
\]
Therefore, the correct option is A.
Additional information:
Few limitations to Raoult’s law are:
1)Raoult's law is useful for describing ideal solutions. However, ideal solutions are rare and hard to find.
2)The negative deviation is seen when the vapour pressure is lower than that of the expected from Raoult's law. A positive deviation is seen when the cohesion between the similar molecules is greater or exceeds the adhesion between dissimilar molecules. Both components of the mixture can easily escape from the solution.
Note: Raoult's law is valid only in ideal solutions. In an ideal solution, the solvent-solute interaction is the same as that of the solvent - solvent or solute - solute interaction. This tells that both the solute and the solvent take the same amount of energy in order to escape to the vapour phase when they exist in their pure states.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

