1)how is the number of moles dissolved in a set amount of solvent related to the molarity? \M = #moles/Liter of solution.
2)how to determine the amount of solvent in 100, 125, 200, and 375 mL of a solution of KI
This isn't clear to me. If the solution is dilute enough, the numbers listed ARE the amounts of solvent. If the solution is not dilute, it would help to have the density of the solution and molality or molarity or percent. (Or we could quantitatively determine the amount of KI there but I doubt that is where you want to do.)
3)how to determine molarity from mass and volume data
mass/molar mass = # moles.
Molarity = #moles/L solution.
we have to make an instrctinal packet showin relationships between moles and concentration.
i cant find the following
1)how is the number of moles dissolved in a set amount of solvent related to the molarity?
2)how to determine the amount of solvent in 100, 125, 200, and 375 mL of a solution of KI
3)how to determine molarity from mass and volume data
websites would be great too thanks...
6 answers
thanks
and for number two i left something out on accident sorry
we had to make a graph of solubility of common inorganic compounds in grams solute per 100 mL solvent
there were several different substances but here is the info for KI
0'C 10'C 20'C 30'C 40'C 50'C
127.5 136 144 152 160 168
'C = degrees celcius
the numbers refer to grams/100gH20
i don't know if this makes it clearer
and for number two i left something out on accident sorry
we had to make a graph of solubility of common inorganic compounds in grams solute per 100 mL solvent
there were several different substances but here is the info for KI
0'C 10'C 20'C 30'C 40'C 50'C
127.5 136 144 152 160 168
'C = degrees celcius
the numbers refer to grams/100gH20
i don't know if this makes it clearer
OK. I know what you're doing now.
For 100 g H2O, just read from the graph at the appropriate T to obtain g KI.
At 125 mL, read at the appropriate T to get g/100, then multiply that by 125/100 or 250/100 or 375/100 or whatever the volume of that part of the problem wants. I hope this helps.
For 100 g H2O, just read from the graph at the appropriate T to obtain g KI.
At 125 mL, read at the appropriate T to get g/100, then multiply that by 125/100 or 250/100 or 375/100 or whatever the volume of that part of the problem wants. I hope this helps.
sorry im not really understanding my graph only goes up to 150g/100H2O so do i have to extend it in order to get the answer
im lost
im lost
If the graph is linear, then you may extrapolate, if you wish, but I don't think that is necessary. Your question originally asked how to determine the solvent but I think you meant to say solute and the last directions I gave were for the solute and not the solvent.
Let's take an example of KI in 100 g H2O at 30 degrees C. Your chart shows 152 g and you said that was in 100 g H2O. So if the question is how much is in 100 g H20, that's it. If the question is how much is in 175 g water, then you do this.
152 g x 175 g/100 g = xx g/175 g.
Or if the question was how much is in 350 g H2O, then the answer is
152 g x 350 g H2O/100 gH2O = yy grams.
I have assumed the density of these solutions is 1.00 which may not be exactly correct but I think you are to equate grams H2O with mL water.
Let's take an example of KI in 100 g H2O at 30 degrees C. Your chart shows 152 g and you said that was in 100 g H2O. So if the question is how much is in 100 g H20, that's it. If the question is how much is in 175 g water, then you do this.
152 g x 175 g/100 g = xx g/175 g.
Or if the question was how much is in 350 g H2O, then the answer is
152 g x 350 g H2O/100 gH2O = yy grams.
I have assumed the density of these solutions is 1.00 which may not be exactly correct but I think you are to equate grams H2O with mL water.
yea i did mean solute sorry
thanks for the help!!
thanks for the help!!
1 answer