Question
What would the freezing point be for a solution of 0.87 M aqueous ammonia? the density of this solution is 0.922 g/cm3.
Answers
bobpursley
Assume you have 922 grams of the stuff, so it has a volume of 1 liter.
Now in that 922 grams, you must have .87 moles of NH3, so figure the mass of that. Say it is 15 grams (it probably isn't, you figure it.). Then you have 907 grams water.
molality=moles/kgwater= .87/.907
freezing point depression= m*(Freezing oint depression for water).
Now in that 922 grams, you must have .87 moles of NH3, so figure the mass of that. Say it is 15 grams (it probably isn't, you figure it.). Then you have 907 grams water.
molality=moles/kgwater= .87/.907
freezing point depression= m*(Freezing oint depression for water).