Question

A \(0.1\) M solution of a solute \(A\) is Isotonic with a solution of \(B\) containing \(1.5\) gm of it per \(100\) cm\({}^{3}\). What is the molar mass of \(B\)?

• A. \(150gmol^{-1}\)
• B. \(115gmol^{-1}\)
• C. \(75gmol^{-1}\)
• D. \(750gmol^{-1}\)

Hint:

Isotonic solutions have equal molar concentration, so equate 0.1 M with the molarity of B.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Isotonic solutions have the same osmotic pressure at the same temperature, so their molarities are equal.

Step 2: Molarity of B
B has 1.5 g in 100 cm\(^3\), which is 15 g per litre. So \(C_B=\dfrac{15}{M_B}\).

Step 3: Equate
\(\dfrac{15}{M_B}=0.1\)

Step 4: Solve
\(M_B=150\,\text{g mol}^{-1}\), option (A).

Final Answer:
The molar mass of B is \(150\,\text{g mol}^{-1}\). \[ \boxed{\text{(A) }150gmol^{-1}} \]