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115 grams of KCl is dissolved in 750 ml of water. what are the molality, molarity, mole percent, % mass, ppm by mass. what woul...Asked by shaknocka lewis
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
chemisty - DrBob222, Monday, February 7, 2011 at 11:30pm
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.
chemisty - Anonymous, Tuesday, February 8, 2011 at 9:34am
are these rite
molality 1.125
molarity-1.2
mole fraction-.34
mole percent- 34%
chemisty - DrBob222, Monday, February 7, 2011 at 11:30pm
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.
chemisty - Anonymous, Tuesday, February 8, 2011 at 9:34am
are these rite
molality 1.125
molarity-1.2
mole fraction-.34
mole percent- 34%
Answers
Answered by
DrBob222
molality. no. I worked the problem for youl. Did you punch in the wrong numbers?
115/74.5 = 1.54 and 1.54/0.750 = aboput 2.06m
Molarity can't be done without the density of the solution.
mole fraction = 0.36
mole % = 36%
115/74.5 = 1.54 and 1.54/0.750 = aboput 2.06m
Molarity can't be done without the density of the solution.
mole fraction = 0.36
mole % = 36%
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