Question

A solution of glucose containing \(18\ \text{g}\) in \(1\ \text{dm}^{3}\) is isotonic with a urea solution. Calculate the mass of urea in \(500\ \text{mL}\) solution.
[Molar mass of urea \(=60\ \text{g mol}^{-1}\). Molar mass of glucose \(=180\ \text{g mol}^{-1}\)]

• A. \(1.5\ \text{g}\).
• B. \(3\ \text{g}\).
• C. \(6\ \text{g}\).
• D. \(9\ \text{g}\).

Hint:

Glucose and urea are non-electrolytes, so isotonic solutions have equal molarity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Both solutes are non-electrolytes, so isotonic solutions have the same molarity.

Step 2: Molarity of glucose
\(\dfrac{18}{180} = 0.1\ \text{mol}\) in \(1\ \text{dm}^3\), so \(C = 0.1\ \text{M}\).

Step 3: Urea in 500 mL
Moles of urea \(= 0.1\times 0.5 = 0.05\ \text{mol}\). Mass \(= 0.05\times 60 = 3\ \text{g}\).

Final Answer:
The urea solution contains 3 g in 500 mL. \[ \boxed{\text{(B) }3\ \text{g}} \]