How to Adjust KNO3 Solution Preparation for Temperature Variations?

AI Thread Summary
To prepare a 500 ml solution of 1 M KNO3 at 20°C using water at 24°C, one must account for the volume changes due to temperature variations. The density of KNO3 and the effects of buoyancy must also be considered in calculations. A formula c1/d1 = c2/d2 can help adjust concentrations and densities for the temperature difference. The discussion highlights confusion over whether to aim for 1 M or 1.000 M KNO3. Ultimately, the problem was solved using the appropriate formulas and corrections for temperature effects.
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Homework Statement



you want to prepare 500 ml of 1 M KNO3 at 20 C, but your lab and water are 24 C. How many grams of solid KNO3 (density= 2.109 g/ml) should be dissolved in a volume of 500 ml at 24 C so that the concentration will be 1 M at 20 C? what apparent mass of KNO3 weighed in air is required?

Homework Equations



Molarity = moles solute/liters solvent

and dimensional analysis


The Attempt at a Solution



i'm really not sure what to do with this..typically i am good with this kind of question but it was kind of thrown at us, and I've never seen the temperature issue before. please help!
 
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Is that really the question? 1 M KNO3? Not 1.000 M?

Your solvent at 24C will occupy a larger volume than at 20C so there is one source of error. That said, your vessel (volumetric flask) will also change its dimensions at 24C vs at 20C that will counteract this tendency. Unless you know both the dV of the vessel and the dV of the solution over the 4C range, you don't know how to solve it analytically. But I digress...

Did the question really ask for 1 M KNO3 or 1.000 M KNO3?

And... will you use the solution at 20C or 24C?
 
it said 1.000 M KNO3. sorry for not specifying.. i was really tired when i typed it out.

i don't know either of these things that you mention. This is only my 2nd day in analytical chem, and this question was given to me as an assignment.

i also think the question is unclear.. but, obviously, there's nothing we can do about it

thanks for your help anyway!
 
i actually solved this problem

the key lies in a formula that was quite difficult to find at first, but quite simple to use afterwards

c1/d1 = c2/d2, where the different concentrations and densities are due to the temperature difference

after that, the buoyancy correction equation is used

thanks for your help anyway
 
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