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
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.