
The vapour pressure of a solution of Urea is 736.2 mm at 100 degree Celsius. Calculate the Osmotic pressure of this solution at 15 degree Celsius.
Answer
503.1k+ views
Hint: Vapour pressure is a colligative property which means that the vapour pressure of solutions is directly proportional to the amount of solute present in a solution. As the amount of solute is increased in the given solvent, the vapour pressure is lowered because fewer solvent molecules are present at the top of the solution which can contribute to the vapour pressure.
Complete answer:
When a non-volatile solute is added into a volatile solvent, the vapour pressure of solution becomes less than the vapour pressure of the pure solvent. Thus we can say that:
Relative lowering of vapour pressure = mole fraction of the solute
This is represented as
\[ \Rightarrow \dfrac{{P_1^0 - {P_1}}}{{P_1^0}} = \dfrac{{{w_2} \times {M_1}}}{{{M_2} \times {w_1}}}\]
Here \[P_1^0\] represents the vapour pressure of the pure solvent
\[{P_1}\] Denotes the vapour pressure after the solute is added
\[{w_2}\] is the weight of solute added
\[{w_1}\] is the weight if solvent
\[{M_1}\] is the molar mass of the solvent
\[{M_2}\] is the molar mass of the solute
\[ \Rightarrow \dfrac{{P_1^0 - {P_1}}}{{P_1^0}}\]
\[ \Rightarrow \dfrac{{760 - 736.2}}{{760}}\]
\[ \Rightarrow 0.03132\]
We can say that this is equal to the molarity times the molar mass of urea
\[ \Rightarrow {\text{Rel}}ative{\text{ }}lowering{\text{ }}in{\text{ }}vapour{\text{ }}pressure = \dfrac{{\pi \times 18}}{{1000 \times 373 \times 0.0821}}\]
\[ \Rightarrow \pi = \dfrac{{72883.454}}{{13860}}\]
\[ \Rightarrow \pi = 41.14atm\]
Therefore we can say that the osmotic pressure of the given solution is 41.14 atm.
Note:
Measurement of osmotic pressure provides a very useful method of determining the molar masses of solutes. This method is very commonly used to determine molar masses of proteins, polymers and other macromolecules which are very unstable at high temperatures. The osmotic pressure method has the advantage that it uses calculations around the room temperature.
Complete answer:
When a non-volatile solute is added into a volatile solvent, the vapour pressure of solution becomes less than the vapour pressure of the pure solvent. Thus we can say that:
Relative lowering of vapour pressure = mole fraction of the solute
This is represented as
\[ \Rightarrow \dfrac{{P_1^0 - {P_1}}}{{P_1^0}} = \dfrac{{{w_2} \times {M_1}}}{{{M_2} \times {w_1}}}\]
Here \[P_1^0\] represents the vapour pressure of the pure solvent
\[{P_1}\] Denotes the vapour pressure after the solute is added
\[{w_2}\] is the weight of solute added
\[{w_1}\] is the weight if solvent
\[{M_1}\] is the molar mass of the solvent
\[{M_2}\] is the molar mass of the solute
\[ \Rightarrow \dfrac{{P_1^0 - {P_1}}}{{P_1^0}}\]
\[ \Rightarrow \dfrac{{760 - 736.2}}{{760}}\]
\[ \Rightarrow 0.03132\]
We can say that this is equal to the molarity times the molar mass of urea
\[ \Rightarrow {\text{Rel}}ative{\text{ }}lowering{\text{ }}in{\text{ }}vapour{\text{ }}pressure = \dfrac{{\pi \times 18}}{{1000 \times 373 \times 0.0821}}\]
\[ \Rightarrow \pi = \dfrac{{72883.454}}{{13860}}\]
\[ \Rightarrow \pi = 41.14atm\]
Therefore we can say that the osmotic pressure of the given solution is 41.14 atm.
Note:
Measurement of osmotic pressure provides a very useful method of determining the molar masses of solutes. This method is very commonly used to determine molar masses of proteins, polymers and other macromolecules which are very unstable at high temperatures. The osmotic pressure method has the advantage that it uses calculations around the room temperature.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

