Question

$0.06\%$ aqueous solution of urea is isotonic with:
A. $0.06\%$ glucose solution
B. $0.6\%$ glucose solution
C. $0.01{\text{M}}$ glucose solution
D. $0.1{\text{M}}$ glucose solution

Answer 1

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.