Question
115 grams of KCl is dissolved in 750 ml of water. what are the molality, molarity, mole percent, % mass, ppm by mass. what would the freezing point and boiling point of that solution assuming Kf of water is 1.86 degree celsius/m and Kb is .0512 degree celsius/m
Answers
I'll start you off with the first one.
molality = moles/kg solvent.
moles = grams/molar mass
kg solvent = 0.750 kg
molar mass KCl about 74.5
Therefore, moles KCl = 115/74.5 = x moles
molality = x moles KCl/0.750 kg. = zz molality.
I'll be glad to help you do the remainder of the problem but I won't work each one for you.
molality = moles/kg solvent.
moles = grams/molar mass
kg solvent = 0.750 kg
molar mass KCl about 74.5
Therefore, moles KCl = 115/74.5 = x moles
molality = x moles KCl/0.750 kg. = zz molality.
I'll be glad to help you do the remainder of the problem but I won't work each one for you.
are these rite
molality 1.125
molarity-1.2
mole fraction-.34
mole percent- 34%
molality 1.125
molarity-1.2
mole fraction-.34
mole percent- 34%
Check your 2-7-11,10:46am post.
Check your 2-7-11,10:51 post