How to Adjust KNO3 Solution Preparation for Temperature Variations?

[tigerlili]

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!

 

[chemisttree]

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?

 

[tigerlili]

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!

 

[tigerlili]

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

 

[DeeNos]

As a scientist, it is important to approach problems systematically and logically. In this case, we can break down the problem into smaller steps to find the solution.

First, we need to calculate the moles of KNO3 needed to make a 1 M solution. This can be done using the given volume (500 ml) and molarity (1 M) using the formula: moles = molarity x volume. This gives us 0.5 moles of KNO3.

Next, we need to consider the temperature difference between the lab temperature (24 C) and the desired temperature (20 C). We can use the equation: M1V1 = M2V2, where M1 and V1 represent the initial molarity and volume, and M2 and V2 represent the final molarity and volume. In this case, we can set up the equation as: (1 M)(500 ml) = (M2)(500 ml), where M2 is the final molarity at 20 C. Solving for M2 gives us a final molarity of 1.2 M.

Now, we can use the formula for molarity to calculate the new mass of KNO3 needed at 24 C. Rearranging the equation, we get mass = moles x molar mass. Using the molar mass of KNO3 (101.1 g/mol), we can calculate the mass of KNO3 needed at 24 C to be 60.6 grams.

Finally, to find the apparent mass of KNO3 weighed in air, we need to consider the density of KNO3 (2.109 g/ml). The apparent mass is the actual mass of the substance plus the mass of the displaced liquid. In this case, the displaced liquid is water, which has a density of 1 g/ml. Therefore, the apparent mass of KNO3 weighed in air would be 60.6 grams + (500 ml x 1 g/ml) = 560.6 grams.

In summary, to prepare 500 ml of 1 M KNO3 at 20 C, we would need to dissolve 60.6 grams of KNO3 at 24 C, with an apparent mass of 560.6 grams. It is important to take into account the temperature difference when preparing solutions, as it can affect the final concentration. I hope this helps and please let me know if