Answer
Verified
399.9k+ views
Hint: Osmotic pressure is directly proportional to concentration and number of molecules. Isotonic solutions are the solutions having the same osmotic pressure and solute concentration when separated by a membrane. The solution depends upon Boyle-van’t Hoff law and Boyle-Charle’s law.
Complete step by step solution:
Colligative properties depend only on the number of dissolved particles in the solution, not on their identity.
Osmosis is a process in which the solvent moves from a region of low concentration of solute to that of high concentration through a semipermeable membrane. Osmotic pressure is the external pressure applied to the solution to prevent it being diluted by the entry of solvent through osmosis.
Osmotic pressure can be calculated by combining Boyle-van’t Hoff law and van’t Hoff-Charle’s law.
i.e.
$\Pi \propto{\text{c}}$
$\Pi \propto {\text{T}}$
\[\Pi \to \] osmotic pressure
\[{\text{c}} \to \] concentration of solute
\[{\text{T}} \to \] absolute temperature
\[\therefore \Pi = {\text{icRT}}\] is the equation to find the osmotic pressure.
\[{\text{i}} \to \]van’t Hoff factor
\[{\text{R}} \to \] gas constant
\[{\text{c}} \to \] concentration of solute
Van’t Hoff factor, \[{\text{i}}\] is the ratio of number of molecules after dissociation to the number of molecules before dissociation.
\[{\text{i}} = \dfrac{{{{\text{N}}_{{\text{after}}}}}}{{{{\text{N}}_{{\text{before}}}}}}\]
For non-electrolytes, \[{\text{i}} = 1\], \[{{\text{N}}_{{\text{after}}}} = {{\text{N}}_{{\text{before}}}}\]
Glucose is an example of non-electrolyte.
Therefore, For glucose, \[{{\text{i}}_{{\text{glucose}}}} = 1\] since it dissociates into two ions.
For urea, \[{{\text{i}}_{{\text{urea}}}} = 1\] since urea does not dissociate into ions.
Mass by volume percentage of urea=$\dfrac{{\text{w}}}{{\text{v}}} = 0.06\% = \dfrac{{0.06}}{{100}}$
$0.06\% \dfrac{{\text{w}}}{{\text{v}}}$aqueous solution of urea means \[\dfrac{{0.06}}{{100}} = 0.06{\text{g}}\] in \[100{\text{mL}}\] or \[0.6{\text{g}}\] in \[1000{\text{mL}}\]
i.e., it contains \[0.6{\text{g}}\] of urea in \[1000{\text{mL}}\] solution or $1{\text{L}}$ of solution.
${\text{c}} = \dfrac{{{\text{wt}}}}{{{\text{mol}}.{\text{mass}} \times {\text{V}}}} = \dfrac{{{\text{0}}{\text{.6g}}}}{{60{\text{g}}.{\text{mo}}{{\text{l}}^{ - 1}} \times 1{\text{L}}}} = 0.01{\text{M}}$
We have to show that which of the concentrations of sodium chloride show the same osmotic pressure as that of urea. Therefore we assume that osmotic pressure of urea is equal to that of glucose
i.e. \[{\Pi _{{\text{urea}}}} = {\Pi _{{\text{glucose}}}}\]
Substituting the equation of osmotic pressure in the above equation,
\[{{\text{i}}_{{\text{urea}}}} \times {{\text{c}}_{{\text{urea}}}} \times {\text{RT}} = {{\text{i}}_{{\text{glucose}}}} \times {{\text{c}}_{{\text{glucose}}}} \times {\text{RT}}\]
\[\because {\text{RT}}\]is equal on both sides, they get cancelled. Hence the equation becomes,
\[{{\text{i}}_{{\text{urea}}}} \times {{\text{c}}_{{\text{urea}}}} = {{\text{i}}_{{\text{glucose}}}} \times {{\text{c}}_{{\text{glucose}}}}\]
Substituting the values, we get
\[1 \times 0.01{\text{M}} = 1 \times {{\text{c}}_{{\text{glucose}}}}\]
\[{{\text{c}}_{{\text{glucose}}}} = 0.01{\text{M}}\]
Therefore concentration of glucose is $0.01{\text{M}}$
Hence option C is correct.
Additional information:
Solutions are of four types-isotonic, hypertonic, iso-osmotic and hypotonic solutions. Hypertonic solution is one which has more concentration than the reference solution. Hypotonic solution is the one which has less concentration than the reference solution. Iso-osmotic solutions are the solutions having the same osmotic pressure but the concentration may or may not be the same.
Note: Osmotic pressure of two solutions having the same molal concentrations are identical. Osmotic pressure stops the flow of solvent with the help of semi-permeable membranes. The Van't Hoff factor is also known as the dissociation factor.
Complete step by step solution:
Colligative properties depend only on the number of dissolved particles in the solution, not on their identity.
Osmosis is a process in which the solvent moves from a region of low concentration of solute to that of high concentration through a semipermeable membrane. Osmotic pressure is the external pressure applied to the solution to prevent it being diluted by the entry of solvent through osmosis.
Osmotic pressure can be calculated by combining Boyle-van’t Hoff law and van’t Hoff-Charle’s law.
i.e.
$\Pi \propto{\text{c}}$
$\Pi \propto {\text{T}}$
\[\Pi \to \] osmotic pressure
\[{\text{c}} \to \] concentration of solute
\[{\text{T}} \to \] absolute temperature
\[\therefore \Pi = {\text{icRT}}\] is the equation to find the osmotic pressure.
\[{\text{i}} \to \]van’t Hoff factor
\[{\text{R}} \to \] gas constant
\[{\text{c}} \to \] concentration of solute
Van’t Hoff factor, \[{\text{i}}\] is the ratio of number of molecules after dissociation to the number of molecules before dissociation.
\[{\text{i}} = \dfrac{{{{\text{N}}_{{\text{after}}}}}}{{{{\text{N}}_{{\text{before}}}}}}\]
For non-electrolytes, \[{\text{i}} = 1\], \[{{\text{N}}_{{\text{after}}}} = {{\text{N}}_{{\text{before}}}}\]
Glucose is an example of non-electrolyte.
Therefore, For glucose, \[{{\text{i}}_{{\text{glucose}}}} = 1\] since it dissociates into two ions.
For urea, \[{{\text{i}}_{{\text{urea}}}} = 1\] since urea does not dissociate into ions.
Mass by volume percentage of urea=$\dfrac{{\text{w}}}{{\text{v}}} = 0.06\% = \dfrac{{0.06}}{{100}}$
$0.06\% \dfrac{{\text{w}}}{{\text{v}}}$aqueous solution of urea means \[\dfrac{{0.06}}{{100}} = 0.06{\text{g}}\] in \[100{\text{mL}}\] or \[0.6{\text{g}}\] in \[1000{\text{mL}}\]
i.e., it contains \[0.6{\text{g}}\] of urea in \[1000{\text{mL}}\] solution or $1{\text{L}}$ of solution.
${\text{c}} = \dfrac{{{\text{wt}}}}{{{\text{mol}}.{\text{mass}} \times {\text{V}}}} = \dfrac{{{\text{0}}{\text{.6g}}}}{{60{\text{g}}.{\text{mo}}{{\text{l}}^{ - 1}} \times 1{\text{L}}}} = 0.01{\text{M}}$
We have to show that which of the concentrations of sodium chloride show the same osmotic pressure as that of urea. Therefore we assume that osmotic pressure of urea is equal to that of glucose
i.e. \[{\Pi _{{\text{urea}}}} = {\Pi _{{\text{glucose}}}}\]
Substituting the equation of osmotic pressure in the above equation,
\[{{\text{i}}_{{\text{urea}}}} \times {{\text{c}}_{{\text{urea}}}} \times {\text{RT}} = {{\text{i}}_{{\text{glucose}}}} \times {{\text{c}}_{{\text{glucose}}}} \times {\text{RT}}\]
\[\because {\text{RT}}\]is equal on both sides, they get cancelled. Hence the equation becomes,
\[{{\text{i}}_{{\text{urea}}}} \times {{\text{c}}_{{\text{urea}}}} = {{\text{i}}_{{\text{glucose}}}} \times {{\text{c}}_{{\text{glucose}}}}\]
Substituting the values, we get
\[1 \times 0.01{\text{M}} = 1 \times {{\text{c}}_{{\text{glucose}}}}\]
\[{{\text{c}}_{{\text{glucose}}}} = 0.01{\text{M}}\]
Therefore concentration of glucose is $0.01{\text{M}}$
Hence option C is correct.
Additional information:
Solutions are of four types-isotonic, hypertonic, iso-osmotic and hypotonic solutions. Hypertonic solution is one which has more concentration than the reference solution. Hypotonic solution is the one which has less concentration than the reference solution. Iso-osmotic solutions are the solutions having the same osmotic pressure but the concentration may or may not be the same.
Note: Osmotic pressure of two solutions having the same molal concentrations are identical. Osmotic pressure stops the flow of solvent with the help of semi-permeable membranes. The Van't Hoff factor is also known as the dissociation factor.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE