The tutor suggests a correction to example 11.2 in Chemistry (sixth edition) by Charles E. Mortimer.

Example 11.2 (page 275) of said textbook is about the Clausius-Clapeyron equation. At one point, the equation becomes

log(0.526/p₁) = 0.491

Then, the antilog is taken; the next line should read

0.526/p₁ = 3.10

From there, what follows agrees with the text:

p₁ = 0.170

HTH:)

Source:

Mortimer, Charles E. Chemistry, sixth edition. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.