Question
Calculate the molarity and the molality of an NH3 solution made up of 30 g of NH3 in 70 g of water. The density of the solution is 0.982 g/ml.
Molar Mass of NH3 = 17.04 g
Molar Mass of NH3 = 17.04 g
Answers
I think I got the answer. Can some please check it.
Molarity = 17.3 mol/L = 17.3 M
Molality = 25.1 mol/kg = 25.1 m
Molarity = 17.3 mol/L = 17.3 M
Molality = 25.1 mol/kg = 25.1 m
I confirmed 25.1 m but not 17.3 M.
moles NH3 = 30/17.04 = ??
volume = mass/density = 70/0.982 = xx and that divided by 1000 to convert to L = zz.
Then M = ??moles/zzL
moles NH3 = 30/17.04 = ??
volume = mass/density = 70/0.982 = xx and that divided by 1000 to convert to L = zz.
Then M = ??moles/zzL
20.1
20.1
The solution of 30% (w/w) NH3, i.e. 30 g NH3 + 70 g H2O = 100 g.
with d = 0.982 g/ml -> 100 g solution = 101.8 ml; and molality = 30 g (NH3)/17 (g/mol) = 1.76 mol of NH3 in 100 g (or 101.8 ml) solution; So Cm = 1.76 (mol)*1000 (ml)/101.8 (ml) = 17.3 mol/L
with d = 0.982 g/ml -> 100 g solution = 101.8 ml; and molality = 30 g (NH3)/17 (g/mol) = 1.76 mol of NH3 in 100 g (or 101.8 ml) solution; So Cm = 1.76 (mol)*1000 (ml)/101.8 (ml) = 17.3 mol/L
why are we multiplying by 1000
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