Question:
Which of the following pair of solutions is isotonic?
\[
A.\ 18\,g/L\ \text{of glucose solution and }6\,g/L\ \text{of urea solution}
\]
\[
B.\ 10\,g/L\ \text{of glucose solution and }10\,g/L\ \text{of urea solution}
\]
\[
C.\ 0.01\,M\ NaOH\ \text{solution and }0.02\,M\ \text{glucose solution}
\]
\[
D.\ 0.01\,M\ NaCl\ \text{solution and }0.01\,M\ \text{glucose solution}
\]
(Assume that NaCl undergoes complete dissociation)
Which of the following pair of solutions is isotonic?
\[ A.\ 18\,g/L\ \text{of glucose solution and }6\,g/L\ \text{of urea solution} \] \[ B.\ 10\,g/L\ \text{of glucose solution and }10\,g/L\ \text{of urea solution} \] \[ C.\ 0.01\,M\ NaOH\ \text{solution and }0.02\,M\ \text{glucose solution} \] \[ D.\ 0.01\,M\ NaCl\ \text{solution and }0.01\,M\ \text{glucose solution} \] (Assume that NaCl undergoes complete dissociation)
\[ A.\ 18\,g/L\ \text{of glucose solution and }6\,g/L\ \text{of urea solution} \] \[ B.\ 10\,g/L\ \text{of glucose solution and }10\,g/L\ \text{of urea solution} \] \[ C.\ 0.01\,M\ NaOH\ \text{solution and }0.02\,M\ \text{glucose solution} \] \[ D.\ 0.01\,M\ NaCl\ \text{solution and }0.01\,M\ \text{glucose solution} \] (Assume that NaCl undergoes complete dissociation)
Show Hint
For isotonic solutions,
\[
iC=\text{same}
\]
Non-electrolytes have
\[
i=1
\]
while electrolytes have higher \(i\) values because of dissociation.
Updated On: Jun 24, 2026
- A and B
- A and C
- B and D
- B and C
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Recall the condition for isotonic solutions.
Two solutions are isotonic if they have the same osmotic pressure.
Osmotic pressure is given by \[ \pi=iCRT \] where \[ i=\text{van’t Hoff factor} \] \[ C=\text{molar concentration} \] At the same temperature, \[ iC \] must be equal for isotonic solutions.
Step 2: Check pair A.
For glucose: \[ \text{Molar mass of glucose}=180\,g/mol \] \[ \frac{18}{180}=0.1\,M \] Glucose is non-electrolyte, so \[ i=1 \] Effective concentration: \[ iC=1\times 0.1=0.1 \] For urea: \[ \text{Molar mass of urea}=60\,g/mol \] \[ \frac{6}{60}=0.1\,M \] Urea is also non-electrolyte, so \[ i=1 \] Effective concentration: \[ iC=1\times 0.1=0.1 \] Hence, pair A is isotonic.
Step 3: Check pair B.
For glucose: \[ \frac{10}{180}=0.0556\,M \] For urea: \[ \frac{10}{60}=0.1667\,M \] Their concentrations are different, so pair B is not isotonic.
Step 4: Check pair C.
NaOH dissociates completely: \[ NaOH \rightarrow Na^+ + OH^- \] Thus, \[ i=2 \] Effective concentration: \[ iC=2\times 0.01=0.02 \] For glucose: \[ i=1 \] Effective concentration: \[ 1\times 0.02=0.02 \] Hence, pair C is isotonic.
Step 5: Check pair D.
NaCl completely dissociates: \[ NaCl \rightarrow Na^+ + Cl^- \] Thus, \[ i=2 \] Effective concentration: \[ 2\times 0.01=0.02 \] For glucose: \[ 1\times 0.01=0.01 \] These are not equal. Hence, pair D is not isotonic.
Step 6: Final conclusion.
Therefore, the isotonic pairs are \[ \boxed{\text{A and C}} \]
Two solutions are isotonic if they have the same osmotic pressure.
Osmotic pressure is given by \[ \pi=iCRT \] where \[ i=\text{van’t Hoff factor} \] \[ C=\text{molar concentration} \] At the same temperature, \[ iC \] must be equal for isotonic solutions.
Step 2: Check pair A.
For glucose: \[ \text{Molar mass of glucose}=180\,g/mol \] \[ \frac{18}{180}=0.1\,M \] Glucose is non-electrolyte, so \[ i=1 \] Effective concentration: \[ iC=1\times 0.1=0.1 \] For urea: \[ \text{Molar mass of urea}=60\,g/mol \] \[ \frac{6}{60}=0.1\,M \] Urea is also non-electrolyte, so \[ i=1 \] Effective concentration: \[ iC=1\times 0.1=0.1 \] Hence, pair A is isotonic.
Step 3: Check pair B.
For glucose: \[ \frac{10}{180}=0.0556\,M \] For urea: \[ \frac{10}{60}=0.1667\,M \] Their concentrations are different, so pair B is not isotonic.
Step 4: Check pair C.
NaOH dissociates completely: \[ NaOH \rightarrow Na^+ + OH^- \] Thus, \[ i=2 \] Effective concentration: \[ iC=2\times 0.01=0.02 \] For glucose: \[ i=1 \] Effective concentration: \[ 1\times 0.02=0.02 \] Hence, pair C is isotonic.
Step 5: Check pair D.
NaCl completely dissociates: \[ NaCl \rightarrow Na^+ + Cl^- \] Thus, \[ i=2 \] Effective concentration: \[ 2\times 0.01=0.02 \] For glucose: \[ 1\times 0.01=0.01 \] These are not equal. Hence, pair D is not isotonic.
Step 6: Final conclusion.
Therefore, the isotonic pairs are \[ \boxed{\text{A and C}} \]
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