Converting units

[KevinE]

Hi guys doing a bit of studying on conversions.


You have 53cm^3 of a 1.5 mol/dm^3 solution of NaCI

But a 0.8 mol/dm^3 solution is needed.

Calculate the volume needed in DM^3


The answer is 0.099 dm^3 apparently???

 

[stapel]

You have 53cm^3 of a 1.5 mol/dm^3 solution of NaCI

But a 0.8 mol/dm^3 solution is needed.

Calculate the volume needed in DM^3

The volume of what? Water (alcohol, whatever solvent) or salt (the solute)?

The answer is 0.099 dm^3 apparently???

What method have they taught you for this? Are you supposed to use the (simplistic, and not valid in "real life") method often taught in beginning algebra course, or are you supposed to use the (real-world) method using molarity (such as here and here)?

When you reply, please show your work so far. Thank you!

 

[KevinE]

I've tried it a few ways - the last was something along these lines

53cm^3 = 148877cm3

1.5mol dm^3 = 3.375 dm3

0.8 dm^3 = 0.512dm3


3.375 / 0.512 = 6.59


Therefore - 148877 / 6.59 = 22591.35 dm3

22591.35 dm3 / 10 = 2259.1 cm3

 

[KevinE]

This is probably some mad crazy thing I've invented myself.

 

[KevinE]

The teachers have gone Denis. It's all me, poor old me left to fend for myself, in the wild mathematically jungle of converia.

I'll be honest and say I don't understand your table. looks fine on the left - 20 + x.
So how could it be 15 on the other? Add it, multiply it, Divide or subtract that - it won't give you 15. So I'm not seeing the pattern.

I have an exam on Tuesday.

 

[stapel]

This is relatively basic stuff; make sure you "follow" it...good luck!

This is relatively basic IF one is doing the simplistic algebra stuff (or liquids, such as in your example). If one is working with moles of actual solute, then the computations require working with the molarity of the substances. :shock:

 

[KevinE]

Thanks Denis I'll have a longer look at it later. It's just above my understanding at the moment. But I'm not so untaught that I can't grasp at it. Cheers guys