## Help With Concentrations Homework

Alain12345 - 2-12-2005 at 18:27

This is probably a stupid question, but I'm still new to chemistry, and I need help. I missed a couple of days at school, and now I'm behind.

I need help with one of the questions I was assigned.

Bags of a D5W intravenous srugar solution used in hospitals contain a 5.0% W/V dextrose-in-water solution.
a) What mass of dextrose is present in a 500.0- mL bag?
b) What is the concentration of D5W expressed in parts per million?

I tried part a, and I assumed you get the answer like this:

C= [m(Solute)/ v (solution)] X 100

5.0%= x/ 500mL

5.0% X 500mL= 2500 g (2.5 kg)

I'm clueless on part b, though.

stygian - 2-12-2005 at 18:40

5 percent?

5 per cent

5 / 100

use .05 not 5

I think. I havent done stoich in so long. Or even any math.

Alain12345 - 2-12-2005 at 18:48

lol I'm so stupid...

Any help with the second part?

Thanks.

EDIT:

Sorry, another question.

The maximum concentration of salt in water at 0*C is 31.6g/100mL. What mass of salt can be dissolved in 250mL of solution?

I just multiplied 31.6g by 2.5 and got 79g. Is this correct?

[Edited on 3-12-2005 by Alain12345]

### well lets see..

stygian - 2-12-2005 at 19:00

We have 5/100. Five parts per hundred. To make hundred a million you take it *10000 if im not mistaken. Then do the same with the 5. Someone correct me if im wrong.
Alain12345 - 2-12-2005 at 20:59

Sorry for all the questions, but I've got one more.

Acid rain may have 355 ppm of dissolved carbon dioxide, which contributes to its acidity.
a) What mass of carbon dioxide is present in 1.00 L of acid rain?
b) Calculate the molar concentration of carbon dioxide in the acid rain sample.

Thanks again.

stygian - 2-12-2005 at 21:18

No guarantees here, but ill give it a shot. Someone correct me if im wrong, then i'll learn something too

maybe im too tired or maybe i just totally suck at everything i used to be good at.

this should be piss easy and ive been stuck for 20 minutes at least. hopefully someone else answers, cuz then i could learn (again)

BromicAcid - 2-12-2005 at 21:18

Parts per million can also be expressed as milligrams per liter (as I recall) using that relation the problem is very simple.
stygian - 2-12-2005 at 21:23

Oh. Then its 355 mg which is..

12(C) + 2*16(O) = 44 (dontremembertheunits)

.355/44 = .008 mol CO2

Damn i gotta brush up on this stuff.

Alain12345 - 2-12-2005 at 23:23

Sorry to bother you guys, but I've got another question... I have to understand this by the end of this weekend...

Phosphoric acid is the active ingredient in many commericial rust-removing solutions. Calculate the volume of concentrated phosphoric acid (14.6 mol/L) that must be diluted to prepare 500mL of a 1.25 mol/L solution.

praseodym - 2-12-2005 at 23:33

(0.5L x 1.25mol/L) / (14.6mol/L) = 0.0428L

Pls correct me if I am wrong.