What is the molality when 0.15 mol SO2 is dissolved in 100 ml water? Remember 1 ml of water = 1 gram of water)

1 Answer
May 21, 2018

"Molality"-=1.5*mol*kg^-1...

Explanation:

By definition, "molality"-="moles of solute"/"kilograms of solvent"

And so here...m_(SO_2)=(0.15*mol)/(100*mLxx1*g*mL^-1xx10^-3kg*g^-1)

-=1.50*mol*kg^-1...

...sometimes this solution is referred to as "sulfurous acid", i.e. H_2SO_3-=H_2O*SO_2...the smell is sulfurous but quite pleasant. Home brewers often use sulfurous acid to sanitize their carboys...