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The boiling point of water at 735mmHg is 99.073C. What mass of NaCl should be added to 3.11kg water to increase the boiling poi...Question
The boiling point of water at 735mmHg is 99.073°C. What mass of BaCl2 should be added to 2.28kg water to increase the boiling point to 100.000°C ?
Kb for water = 0.510 K kg mol-1
Kb for water = 0.510 K kg mol-1
Answers
DrBob222
delta T = i*Kb*molality..............i for BaCl2 = 3
100.000 - 99.073 = 0.927 = 3*0.510*m
Solve for m.
Then m = moles/kg solvent. You know m and kg solvent. Solve for moles.
Then moles = grams/molar mass. You know moles and molar mass. Solve for grams.
Post your work if you get stuck.
100.000 - 99.073 = 0.927 = 3*0.510*m
Solve for m.
Then m = moles/kg solvent. You know m and kg solvent. Solve for moles.
Then moles = grams/molar mass. You know moles and molar mass. Solve for grams.
Post your work if you get stuck.