(a) \[{\text{200 c}}{{\text{m}}^{\text{3}}}\] of an aqueous solution of a protein contains \[1.26{\text{ g}}\] of protein. The osmotic pressure of such a solution at \[{\text{300 K}}\] is sound to be \[{\text{2}}{\text{.57 }} \times {\text{ 10}}{{\text{ }}^{{\text{ - 3}}}}{\text{ bar}}\] . Calculate molar mass of protein.
(b) Plot a graph of vapour pressure against mole fractions of the two volatile liquids forming an ideal solution
Answer
549.3k+ views
Hint:(a) To determine calculate the molar mass from the osmotic pressure, use the following relationship
\[M = \dfrac{{w \times R \times T}}{{\Pi \times V}}\]
(b) An ideal solution obeys Raoult’s law under all conditions of temperature and pressure.
According to Raoult's Law, the vapour pressure of a solution of non-volatile solute is equal to the product of the pure solvent and the mole fraction of the solvent.
Complete step-by-step solution:
The mass of the protein \[{\text{w = }}1.26{\text{ g}}\]
The volume of the solution \[{\text{V = 200 c}}{{\text{m}}^{\text{3}}} = 0.2{\text{ L}}\]
The temperature of the solution \[{\text{T = 300 K}}\]
The osmotic pressure of the solution \[\Pi = {\text{2}}{\text{.57 }} \times {\text{ 10}}{{\text{ }}^{\text{3}}}{\text{ bar}}\]
Write the formula for the osmotic pressure of the solution
\[\Pi = CRT = \dfrac{{nRT}}{V}\]
\[\Pi = \dfrac{{wRT}}{{MV}}{\text{ }}\left( {\because n = \dfrac{w}{M}} \right)\]
Rearrange the above formula
\[M = \dfrac{{w \times R \times T}}{{\Pi \times V}}\]
Substitute values in the above formula
\[M = \dfrac{{1.26{\text{ g}} \times 0.08206 \times {\text{300 K}}}}{{{\text{2}}{\text{.57 }} \times {\text{ 10}}{{\text{ }}^{{\text{ - 3}}}}{\text{ bar}} \times 0.2{\text{ L}}}}\]
\[M = 61,039{\text{ g/mol}}\]
Hence, the molar mass of the protein is \[61,039{\text{ g/mol}}\] .
(b)Consider a binary mixture of two liquids A and B
\[P_A^o\] and \[P_B^o\] are the vapour pressures of pure liquids A and B respectively.
\[{X_A}\] and \[{X_B}\] are the mole fractions of liquids A and B respectively in the solution.
Plot a graph of vapour pressure against mole fractions of the two volatile liquids forming an ideal solution:
Note:(a)Calculating molar mass from osmotic pressure is more accurate than calculating the molar mass from either the elevation in the boiling point or the depression in the freezing point.
(b) For an ideal binary solution containing two components A and B. the expression for the total pressure is as follows.
\[
{P_T} = {P_A} + {P_B} \\
{P_T} = {X_A}P_A^o + {X_B}P_B^o \\
{P_A} = {X_A}P_A^o \\
{P_B} = {X_B}P_B^o \\
\]
\[M = \dfrac{{w \times R \times T}}{{\Pi \times V}}\]
(b) An ideal solution obeys Raoult’s law under all conditions of temperature and pressure.
According to Raoult's Law, the vapour pressure of a solution of non-volatile solute is equal to the product of the pure solvent and the mole fraction of the solvent.
Complete step-by-step solution:
The mass of the protein \[{\text{w = }}1.26{\text{ g}}\]
The volume of the solution \[{\text{V = 200 c}}{{\text{m}}^{\text{3}}} = 0.2{\text{ L}}\]
The temperature of the solution \[{\text{T = 300 K}}\]
The osmotic pressure of the solution \[\Pi = {\text{2}}{\text{.57 }} \times {\text{ 10}}{{\text{ }}^{\text{3}}}{\text{ bar}}\]
Write the formula for the osmotic pressure of the solution
\[\Pi = CRT = \dfrac{{nRT}}{V}\]
\[\Pi = \dfrac{{wRT}}{{MV}}{\text{ }}\left( {\because n = \dfrac{w}{M}} \right)\]
Rearrange the above formula
\[M = \dfrac{{w \times R \times T}}{{\Pi \times V}}\]
Substitute values in the above formula
\[M = \dfrac{{1.26{\text{ g}} \times 0.08206 \times {\text{300 K}}}}{{{\text{2}}{\text{.57 }} \times {\text{ 10}}{{\text{ }}^{{\text{ - 3}}}}{\text{ bar}} \times 0.2{\text{ L}}}}\]
\[M = 61,039{\text{ g/mol}}\]
Hence, the molar mass of the protein is \[61,039{\text{ g/mol}}\] .
(b)Consider a binary mixture of two liquids A and B
\[P_A^o\] and \[P_B^o\] are the vapour pressures of pure liquids A and B respectively.
\[{X_A}\] and \[{X_B}\] are the mole fractions of liquids A and B respectively in the solution.
Plot a graph of vapour pressure against mole fractions of the two volatile liquids forming an ideal solution:
Note:(a)Calculating molar mass from osmotic pressure is more accurate than calculating the molar mass from either the elevation in the boiling point or the depression in the freezing point.
(b) For an ideal binary solution containing two components A and B. the expression for the total pressure is as follows.
\[
{P_T} = {P_A} + {P_B} \\
{P_T} = {X_A}P_A^o + {X_B}P_B^o \\
{P_A} = {X_A}P_A^o \\
{P_B} = {X_B}P_B^o \\
\]
Recently Updated Pages
Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

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

Explain the structure of megasporangium class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

