Asked by Anon

What is the boiling point elevation of a solution of NaNO3 (85.0 g/mol, complete dissociation) made by dissolving 10.0 g of NaNO3 into 200 g of water (Kb = 0.512◦C/m)?
Answered by DrBob222
delta T = i*Kb*m
i = 2 for NaNO3
Kb is given
m = mols/Kg. Kg is 0.2
mols = grams/molar mass NaNO3.