Chemistry problem.....

[intothevoidx]

I was wondering if there is some kind of equation out there to solve a problem like this. How many milliliters of a 25% solution of potassium chloride must be added to 8 milliliters of pure potassium chloride to obtain a 35% solution.

 

 

 

thanks

[jdurg]

Can't be solved. The reason for this is that potassium chloride is an ionic solid. It is not a liquid so you can not have a pure solution of it. It would be like asking how many milliliters of Frosted Flakes do you need to add to a bowl of milk to get five hundred grams of cereal.

[intothevoidx]

OK but mathematically how do you solve it, if it were possible?

[akcapr]

youd find out the density of kcl when its moleten then figure out the mols and go from there i think

[intothevoidx]

I'm pretty sure there is either an algebraic equation you can set up for a problem like this or a chemistry equation that you can use to solve it.

[NeonBlack]

molten KCl.... HA!

 

If you replace KCl with something like say for example alcohol, think about it like this:

 

You have 8ml of 100% add V ml of 25% solution, and you end up with V+8 ml of 35% solution.

[woelen]

Of course this problem can be solved theoretically, but you have to make an important assumption.

The solid KCl must be completely compact (no air between the granules). Otherwise a statement about ml's of this solid is meaningless. Usually, with solids we talk about weight and your question then can be restated as follows:

 

"I was wondering if there is some kind of equation out there to solve a problem like this. How many milliliters of a 25% solution of potassium chloride must be added to ... grams of pure potassium chloride to obtain a 35% solution."

 

You need to know the density of a 25% solution of potassium chloride. You can measure this, or use some published table.

 

With that info the problem is simple. Suppose you have X grams of KCl, then you proceed as follows:

 

Suppose you have Y grams of a 25% solution of KCl and you add X grams to this and all of X dissolves (assuming that KCl is so soluble that a 35% solution can be made), then you have a liquid with (0.25*Y + X) grams of solution. The total weight of this is Y + X grams. So, you need to solve the following equation:

 

0.25*Y + X = 0.35*(Y+X)

 

Solving this is piece of cake, for a given Y.

[Borek]

You may use bulk density for loose KCl. Engineers do such things every day

 

Best,

Borek

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