Question

A \(6%\) solution of urea is isotonic with:

• A. \(1\ \text{M}\) solution of glucose
• B. \(0.05\ \text{M}\) solution of glucose
• C. \(6%\) solution of glucose
• D. \(25%\) solution of glucose

Hint:

For isotonic solutions of non-electrolytes, equality of molarity is sufficient since their van’t Hoff factor (i = 1).

Answer 1

The Correct Option is A

Step 1: Understand the meaning of isotonic solutions. Two solutions are isotonic if they have the same osmotic pressure at the same temperature. \[ \pi = iMRT \] Step 2: Calculate molarity of the urea solution. A \(6%\) solution means: \[ 6\ \text{g urea in } 100\ \text{mL solution} \] So, in \(1000\ \text{mL}\): \[ = 60\ \text{g urea per litre} \] Molar mass of urea \(= 60\ \text{g mol}^{-1}\) \[ \text{Molarity of urea} = \frac{60}{60} = 1\ \text{M} \] Step 3: Compare van’t Hoff factors. Urea is a non-electrolyte: \[ i_{\text{urea}} = 1 \] Glucose is also a non-electrolyte: \[ i_{\text{glucose}} = 1 \] Step 4: Apply isotonic condition. For isotonic solutions: \[ i_1 M_1 = i_2 M_2 \] \[ (1)(1\ \text{M}) = (1)(M_{\text{glucose}}) \] \[ M_{\text{glucose}} = 1\ \text{M} \] Hence, a \(6%\) solution of urea is isotonic with a \(\boxed{1\ \text{M}}\) solution of glucose.