Dilution Problems

Last updated 10-11-09

Dilution Problem Quiz

You may remember this collage of things that relate to hydrogen peroxide (H₂O₂). The only difference between all of these is the concentration of hydrogen peroxide. The jet pack and rockets use 90% w/v. Bleaching of the hair uses between 3% w/v and 6% w/v. The hydrogen peroxide put on cuts is 3% w/v. Labs often keep 30% w/v solutions around.
Problem 1: How many mL of 30% w/v hydrogen peroxide would you need to make 550mL of 3% w/v hydrogen peroxide (H3)?
(Hint 550mL and 3% are the two knowns given)

	A	B	C	D	E	F	G	H	I
1	Given concentration and volume (weaker)				Stronger inverted
2	3% w/v				Stronger inverted
3	3	g H₂O₂	550	mL of weaker	100	mL of stronger	=	??	mL
4	100	mL of weaker			30	g H₂O₂

Here is a picture of hands that had contacted 90% w/v hydrogen peroxide. Again, it's dangerous unless diluted to about 3% w/v. Then you can even use it as a mouthwash. Problem 2: Someone took 10mL of 90% w/v hydrogen peroxide and diluted it by bringing the volume to 250mL, what is the final concentration in % w/v?
(This time the given was the stronger solution. Also instead of asking for volume of weaker, it's asking for the concentration. Don't include the 100 at G4 in calculation, that 100 stays and becomes I4)

	A	B	C	D	E	F	G	H	I	J
1	Given concentration times volume to get grams H₂O₂				weaker mL placed to have vol in denominator		makes answer per 100mL		Final concentration after dilution
2	90% w/v		vol of stronger		weaker mL placed to have vol in denominator		makes answer per 100mL		Final concentration after dilution
3	90	g H₂O₂	10	mL of stronger			100	=	??	g
4	100	mL of stronger			250	mL of weaker	100		100	mL wealer

Problem 3: What is the formula put into I3? (Hint: It starts as =A3*C3...)

Here is a picture of someone handling 30% hydrogen peroxide safely with gloves. The manufacturer of this bottle listed it as 30 wt%, which is another way of writing 30% w/w (30g H₂O₂ per 100g of solution). The person poured out 300mL but then had to weigh it because concentration is 30g H₂O₂ per 100g (not 100mL). The weight was 330grams. The person added 2 liters (2 kilograms) of water to dilute it. Problem 4: What is the final concentration in % w/w (I3)? (Hint: Don't use the 100 in G4 in calculation. It stays and becomes I4).

	A	B	C	D	E	F	G	H	I	J
1	Given concentration times grams of solution to get grams H₂O₂				total grams of solution after dilution put in denominator		makes answer per 100g		Final concentration after dilution in %w/w
2	30% w/w		mass of stronger		total grams of solution after dilution put in denominator		makes answer per 100g		Final concentration after dilution in %w/w
3	30	g H₂O₂	330	g strong solution			100	=	??	g H₂O₂
4	100	g strong solution			2330	g weak solution	100		100	g of weak

Here's a 1000mL bottle of hydrochloric acid solution. It's hard to read but it says "6 molar" below the name. Under the manufacturer name it says, "Your safer source of science supplies". In a way, that's correct. Normally, concentrated hydrochloric acid is 12 molar and comes in glass bottles. This is half that strength and comes in non-breakable plastic bottles. Stomach acid is 0.1 molar. Problem 5: To what volume (mL) would you need to make it so that 170mL of this 6 molar HCl would become 0.1 molar HCl?

	A	B	C	D	E	F	G	H	I
1	moles per liter times mL gives millimoles HCl				weaker solution conc. inverted to cancel moles		Final mL of weaker solution
2	6 molar HCl		volume of 6M HCl		weaker solution conc. inverted to cancel moles		Final mL of weaker solution
3	6	moles HCl	170	mL of strong solution	?	Liter	=	???	mL
4	1	Liter of strong solution			?	mole HCl

To the left is an IV bag that shows 5% Dextrose and 0.9% sodium chloride. Let's say they got a batch of these IV bags but it didn't have the 0.9% sodium chloride. So the hospital wanted the nurses to use a 20 mL syringe to inject 20 mL of a concentrated sodium chloride solution into the IV bag so that would give the 500mL bag a 0.9% w/v sodium chloride concentration. Problem 6: What is the NaCl concentration needed in the syringe (I3) which will be in % w/v ? (Hint: notice I used 520mL as the diluted volume because of the extra mL added. Also, don't forget to put in the proper number of mL in A4 before calculating.)

	A	B	C	D	E	F	G	H	I	J
1	Concentration times volume gives grams NaCl				1mL of strong solution in denominator to end with "per 100 mL"		makes answer per 100mL		Final concentration after dilution in %w/w
2	0.9% NaCl		volume of IV bag				makes answer per 100mL		Final concentration after dilution in %w/w
3	0.9	g NaCl	520	mL final solution			100	=	??	g NaCl
4	???	mL weak solution			20	mL of strong solution	100		100	mL of strong

Problem 7: What is the formula put in I3?

To the left is a micropipette. The label says it delivers 500 microliters (µl or µL). Let's say the liquid you delivered was 8.5M (moles per liter) mercury (II) chloride. You added that to a beaker that contained a certain number of mL of water. That number of mL added will be the number that matches the atomic number of your element. For example, if your element code name is sulfur, which has atomic number of 16, you would assume 16mL of water. Problem 8: What is the final concentration of mercury (II) chloride in mg per mL? (Hint: See spreadsheet below to help get started). As usual we are given a concentration and a volume that goes together. So that starts the dimensional analysis. Since the answer wants "per mL" we put the mL of water in the denominator. Concentration is in moles per liter, so you will have to change moles to grams because the answer is in mg per mL. Of course, the 8.5M is automatically turned in to 8.5 moles per liter.

moles per liter times liter makes moles

cancel µ

Use your atomic # as mL of water, which is final volume

mole-to-gram with molar mass

add milli

Final mg/mL of diluted solution

Concentration of strong solution 8.5M

vol. strong

???

moles of HgCl₂

500

µL

10^-6

???

g. HgCl₂

???

Liter of strong solution

???

mL of diluted solution

mole HgCl₂

???

Note at column E, you have got moles of HgCl₂. At column G, it is putting those moles over mL of water. So you are basically showing moles per liter of the diluted solution. At column I, this is now grams per mL of solution. At column L, you have milligrams of HgCl₂ per milliliter of diluted solution. You have to plug in values for A3, F4, H3, and J4 before doing the final calculation. Remember your answer will be different than anyone else's answer.

CHM151 students, send your answers to chm151@chemistryland.com. In subject line use: Dilution Quiz.

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Since Oct., 2009