Counting molecules in a micro-drop:\nA drop of water weighs 0.018 g. It contains 6.023 × 10^? molecules. What is the missing exponent?

Introduction / Context:
Back-of-the-envelope mole–molecule conversions are invaluable in reaction engineering and analytical chemistry. Here, we estimate the number of molecules in a tiny water drop using molar mass and Avogadro’s number.

Given Data / Assumptions:

• Mass of water drop = 0.018 g.
• Molar mass of water = 18 g/mol.
• Avogadro’s number ≈ 6.023 × 10^23 molecules/mol.

Concept / Approach:
Number of moles n = mass / molar mass. Number of molecules N = n × N_A. Because 0.018 g is a thousandth of 18 g, the number of moles is 0.001 mol, i.e., 10^-3 mol. Multiply by 6.023 × 10^23 to obtain 6.023 × 10^20 molecules.

Step-by-Step Solution:
n = 0.018 g / 18 g·mol^-1 = 0.001 mol.N = n × N_A = 0.001 × 6.023 × 10^23 = 6.023 × 10^20 molecules.Therefore, the missing exponent is 20.

Verification / Alternative check:
Scale reasoning: 1 mol has 6.023 × 10^23 molecules; 10^-3 mol has 10^-3 of that count → exponent 23 − 3 = 20, consistent with the calculation.

Why Other Options Are Wrong:
10^23 corresponds to 1 mol, not 0.001 mol; 10^26 and 10^19 or lower deviate by orders of magnitude; 10^17 is far too small.

Common Pitfalls:
Using 18 kg/kmol incorrectly; forgetting to convert mass to moles; misplacing powers of ten.

Final Answer:
10^20