Question

The number of atoms in 21.6 g of Ag (atomic mass of Ag = 108) is equal to the number of molecules present in:

• A. 1.8 g of H₂O
• B. 12 moles of KMnO₄
• C. 0.6 N H₂SO₄
• D. 4.6 g of C₂H₅OH

Hint:

Convert the Ag mass to moles first, then compare with each option's mole count.

Answer 1

The Correct Option is B

Step 1: Find the moles of Ag atoms in 21.6 g. Moles = mass / molar mass = 21.6 / 108 = 0.2 mol. So the sample contains 0.2 mol of Ag atoms, which is 0.2 x 6.022 x 10^23, about 1.2 x 10^23 atoms.

Step 2: Check option A. 1.8 g of H₂O has molar mass 18, so moles = 1.8/18 = 0.1 mol, which is only half of the 0.2 mol needed.

Step 3: Check option D. 4.6 g of C₂H₅OH has molar mass 46, so moles = 4.6/46 = 0.1 mol, again only half of the required 0.2 mol.

Step 4: Check option C. 0.6 N H₂SO₄ is a statement about concentration, normality, not about a fixed mass or a fixed number of particles, so without a stated volume it cannot be compared directly to a fixed particle count.

Step 5: The key for this question is 12 moles of KMnO₄, option B. Taken literally, 12 moles of KMnO₄ contains far more particles than 0.2 mol of Ag, 12 mol against 0.2 mol, so this option does not line up numerically with the given data. This looks like an inconsistency in the original question or key, most likely a much smaller quantity of KMnO₄ was intended. As per the answer key provided for this paper, option B is marked correct, so B is taken as the answer here.