Question

A solution containing $8.3~g~dm^{-3}$ of Urea (molar mass $=60~g~mol^{-1}$) is found to be isotonic with $5%$ solution of non-volatile organic solute A. The approximate molar mass of A is}

• A. $1348.8~g~mol^{-1}$
• B. $361.5~g~mol^{-1}$
• C. $36.1~g~mol^{-1}$
• D. $34.9~g~mol^{-1}$

Hint:

Logic Tip: Isotonic solutions have equal osmotic pressure, so for non-electrolytes their molar concentrations are equal.

Answer 1

The Correct Option is B

Concept:
Two solutions are said to be isotonic when they have the same osmotic pressure. For dilute solutions of non-electrolytes, osmotic pressure depends upon molarity. Therefore, isotonic solutions contain equal molar concentrations. 

Step 1: 
Given: 
• Mass of urea $= 8.3~g$ 
• Molar mass of urea $= 60~g~mol^{-1}$ 
• Volume $= 1~dm^3$ Number of moles of urea: \[ \text{Moles}=\frac{8.3}{60} \] \[ =0.1383~mol \] Since volume is $1~dm^3$, \[ \text{Molarity}=0.1383~M \] 

Step 2: 
A $5%$ solution means: \[ 5~g \text{ in } 100~mL \] Therefore, in $1000~mL = 1~dm^3$: \[ 50~g \text{ of solute A} \] Let molar mass of A be $M$. \[ \text{Moles of A}=\frac{50}{M} \] Thus, \[ \text{Molarity of A}=\frac{50}{M} \] 

Step 3: 
Since both solutions are isotonic: \[ 0.1383=\frac{50}{M} \] \[ M=\frac{50}{0.1383} \] \[ M \approx 361.5~g~mol^{-1} \] 

Step 4: 
Therefore, the approximate molar mass of solute A is: \[ \boxed{361.5~g~mol^{-1}} \] Hence, the correct answer is: \[ \boxed{\text{(2) }361.5~g~mol^{-1}} \]