Question

\(x\%\;(w/v)\) solution of urea is isotonic with \(4\%\;(w/v)\) solution of a non-volatile solute of molar mass \(120\;g\;mol^{-1}\). The value of \(x\) is

• A. \(2\)
• B. \(4\)
• C. \(3\)
• D. \(5\)

Hint:

For isotonic non-electrolyte solutions, molarity must be equal. In \(w/v\%\), the given mass is always per \(100\;mL\) of solution.

Answer 1

The Correct Option is A

Step 1: Understand the condition of isotonic solutions.
Two solutions are called isotonic when they have the same osmotic pressure.
For dilute solutions, osmotic pressure is given by:
\[ \pi = CRT \] where \(C\) is molar concentration, \(R\) is gas constant, and \(T\) is temperature.
Since both solutions are isotonic at the same temperature, their molar concentrations must be equal.
Therefore:
\[ C_{\text{urea}} = C_{\text{solute}} \]

Step 2: Calculate molarity of the non-volatile solute solution.
A \(4\%\;(w/v)\) solution means:
\[ 4\;g \text{ of solute is present in } 100\;mL \text{ of solution} \] Given molar mass of solute:
\[ 120\;g\;mol^{-1} \] Number of moles of solute in \(100\;mL\):
\[ \text{Moles}=\frac{4}{120} \] \[ =\frac{1}{30} \] Since volume is \(100\;mL = 0.1\;L\), molarity is:
\[ C_{\text{solute}}=\frac{1/30}{0.1} \] \[ =\frac{1}{3}\;M \]

Step 3: Apply isotonic condition to urea solution.
Since urea solution is isotonic with this solution:
\[ C_{\text{urea}}=\frac{1}{3}\;M \]

Step 4: Calculate mass of urea required in \(100\;mL\).
Molar mass of urea is:
\[ 60\;g\;mol^{-1} \] Moles of urea required in \(100\;mL = 0.1\;L\):
\[ \text{Moles of urea}=C \times V \] \[ =\frac{1}{3}\times 0.1 \] \[ =\frac{1}{30}\;mol \] Mass of urea required:
\[ \text{Mass}=\text{Moles}\times \text{Molar mass} \] \[ =\frac{1}{30}\times 60 \] \[ =2\;g \]

Step 5: Convert into percentage \(w/v\).
Since \(2\;g\) urea is present in \(100\;mL\) solution, the percentage strength is:
\[ 2\%\;(w/v) \] Therefore:
\[ x=2 \]

Step 6: Final conclusion.
Hence, the value of \(x\) is:
\[ \boxed{2} \]