Question

3 x 10\(^{22}\) molecules of Na\(_2\)CO\(_3\) (molecular weight = 106) present in 500 ml of solution. The normality of the solution formed is (N = 6 x 10\(^{23}\) mol\(^{-1}\))

• A. 0.1 N
• B. 0.2 N
• C. 0.4 N
• D. 0.05 N

Hint:

The n-factor is crucial for converting between molarity and normality. For acids, it's the number of H⁺ ions; for bases, the number of OH⁻ ions; and for salts, it's the total charge on the cations (or anions).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
3×10²² molecules Na₂CO₃ in 500ml. Molar mass 106. Find normality.

Step 2: Key Formula (Alternate):
N = M × n-factor. For Na₂CO₃, n-factor=2.

Step 3: Detailed Explanation:
Moles=3×10²²/6×10²³=0.05. M=0.05/0.5=0.1. N=0.1×2=0.2N.

Step 4: Final Answer:
Normality is 0.2 N.