Wonder who would offer that as a "sample"?
Molarity: grams/molmass*volumeInLiters
so if you had one liter of this stuff, you so assume 1 liter of this stuff, which is 1100g
.15*1100/98*1=165/98 Moles/liter
Normality= 2*molarity
Molality=.15*1100/98*1.1
A commertially available sample of H2SO4 is 15% of H2SO4 by weight. Density=1.10g. Calculate-
1. Molarity
2. Normality
3. Molality
3 answers
I believe I see a typo here for molality.
grams H2SO4 = 1100*0.15 = 165 as stated.
mols H2SO4 = 0.15*100/98 as stated.
grams solution = 1100 g as stated
BUT g solvent = 1100-165 = 935g (0.935 kg) so
molality = 0.15*1100/0.935 kg solvent = ?
grams H2SO4 = 1100*0.15 = 165 as stated.
mols H2SO4 = 0.15*100/98 as stated.
grams solution = 1100 g as stated
BUT g solvent = 1100-165 = 935g (0.935 kg) so
molality = 0.15*1100/0.935 kg solvent = ?
yes, my eyes are failing.