Water vapour pressure at the boiling point:\nAt 100 °C, the equilibrium vapour pressure of pure water is closest to which of the following values?

Introduction / Context:
By definition, at its normal boiling point (100 °C), water’s vapour pressure equals 1 atmosphere. Converting 1 atm into different pressure units is a frequent exam task and a useful field skill.

Given Data / Assumptions:

• 1 atm ≈ 760 mm Hg ≈ 76 cm Hg ≈ 1.01325 bar.
• Mercury density ≈ 13.6 g/cm^3 (giving the 13.6 factor relating water vs mercury columns).

Concept / Approach:
Match 1 atm to the listed forms. 76 cm of mercury is equivalent to 760 mm of mercury, which equals approximately 1 atm at 0 °C calibration conditions. Therefore, “76 cm of Hg” is the correct choice among the options provided.

Step-by-Step Solution:
Recognize boiling point definition: Pvap = 1 atm.Convert 1 atm to a mercury column: 760 mm Hg = 76 cm Hg.Select “76 cm of Hg.”

Verification / Alternative check:
1.013 bar also equals 1 atm; however it is not one of the offered correct forms here (unless explicitly listed). “76 cm Hg” matches exactly.

Why Other Options Are Wrong:
100 N/m^2 is only 1 Pa, far too small; 13.6 cm Hg is 0.18 atm; 760 mm water column is much less than 1 atm due to water’s lower density; 1.013 bar would be correct numerically but is not the option chosen by the exam’s framing.

Common Pitfalls:
Mixing up water vs mercury manometer columns; remember the density factor 13.6.

Final Answer:
76 cm of Hg