Chem Help! What is the boiling point of water at an elevation of 1.15×10^4 ft?

The atmospheric pressure at sea level is 760 mm Hg. This pressure decreases by 19.8 mm Hg for every 1000- ft increase in elevation.

Answer 1

using your information
1.15E4 ft * 19.8 mmHg/1000 ft * -0.05 C/1 mm = -11.385 C
100 – 11.385 = 88.615 C or 88.6 C

Answer 2

19.8 mmHg/1000 ft times 1.15 x 10^4 feet = 227.7 mmHg

227.7 mmHg times 0.05 C/mmHg = 11.385 C

100 C minus 11.385 C = 88.615 C

Check your math AND my math.

Answer 3

Use the equation: ΔT = mKb ΔT = change in temperature = 0.203C m = molality of solution (unknown) Kb = ebullioscopic constant of water = 0.512 C/m Therefore, 0.203C = m(0.512 C/m) molality of solution = 0.396 moles/kg Note: molality of solution = moles solute/kg of solvent Therefore, 0.396 m = moles solute/0.05 kg solvent moles solute = 0.0198 moles solute Note: molar mass = mass (in grams)/ number of moles molar mass = 4.21 grams/0.0198 moles = 212.6 grams/mole That’s it!n_n

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