Determine the boiling point of water at 698.5 mm Hg.

Use the Clausius-Clapeyron equation. You must look up delta H vaporization. Data not listed you will need: vapor pressure H2O at 100 C = 760 mm Hg.

To determine the boiling point of water at a specific pressure, we can use the Clausius-Clapeyron equation:

ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)

Where:
- P1 and P2 are the initial and final pressures (in this case, the initial pressure is the standard atmospheric pressure of 760 mm Hg, and the final pressure is 698.5 mm Hg).
- ΔHvap is the heat of vaporization of water, which is approximately 40.7 kJ/mol.
- R is the ideal gas constant, which is approximately 8.314 J/(mol·K).
- T1 is the initial temperature of water, which is the boiling point at the initial pressure (760 mm Hg), and T2 is the temperature at the final pressure (698.5 mm Hg).

To solve for T2, we rearrange the equation as follows:

1/T2 = (1/T1) - (R/ΔHvap) * ln(P1/P2)

Now, let's substitute the values into the equation:

T1 (initial boiling point at 760 mm Hg) = 100 degrees Celsius = 373.15 K

P1 (initial pressure) = 760 mm Hg
P2 (final pressure) = 698.5 mm Hg

Now we can plug the values into the equation:

1/T2 = (1/373.15 K) - (8.314 J/(mol·K) / 40.7 kJ/mol) * ln(760 mm Hg / 698.5 mm Hg)

Calculating the right side of the equation:

1/T2 = (1/373.15 K) - (0.203 K/mol) * ln(1.089)

Using a calculator to evaluate the natural logarithm and perform the calculations:

1/T2 ≈ (1/373.15 K) - (0.203 K/mol) * (-0.080)

Adding the values on the right side:

1/T2 ≈ (1/373.15 K) + 0.01624 K

Subtracting the fractions on the right side:

1/T2 ≈ (1/373.15 K) + 0.01624 K / 1

Combining the fractions:

1/T2 ≈ (1 + 0.01624 * 373.15) / 373.15 K

1/T2 ≈ 6.060295 / 373.15 K

Using the reciprocal to solve for T2:

T2 ≈ 373.15 K / 6.060295

T2 ≈ 61.51 degrees Celsius

Therefore, the boiling point of water at 698.5 mm Hg is approximately 61.51 degrees Celsius.