Simplified 2.0 mL .20M X added to 9ml

  • Thread starter Whalstib
  • Start date
In summary: Warren's homework statement was that 2 ml of .20M "X" was added to 9 ml of water to create an 11 mL solution. The concentration of "X" is unknown. Homework equations state that M=molarity=moles/liters. Warren attempted to solve the problem by adding .0004 moles of "X" to .011 liters of water, resulting in a concentration of .0363636...M "X". It was suggested that if his first line was "2.0ml=.002L", then it would be ok; but he instead had "2.0ml=.002L x .2M". 2mL is indeed equivalent to 0.002L,
  • #1
Whalstib
119
0
Simplified... 2.0 mL .20M "X" added to 9ml

Homework Statement



Simplified...

2.0 mL .20M "X" added to 9ml making an 11mL soln.

What would be the concentration of "X"?

Homework Equations



M= molarity=moles/liters

The Attempt at a Solution



2.0ml=.002L x .2M= .0004 moles/.011L=.0363636...M "X"

Right?...

Thanks,

Warren
 
Last edited:
Physics news on Phys.org
  • #2


Whalstib said:

Homework Statement



Simplified...

2.0 mL .20M "X" added to 9ml making an 11mL soln.

What would be the concentration of "X"?


Homework Equations



M= molarity=moles/liters



The Attempt at a Solution



2.0ml=.002L x .2M= .004 moles/.011L=.363636...M "X"

Right?...

Thanks,

Warren

Sure about that? :smile:
 
  • #3


Now I am...

.0363636...
 
  • #4


Whalstib said:
2.0ml=.002L x .2M= .0004 moles/.011L=.0363636...M "X"

It suggests correct thinking, but writing 2 mL=0.0004 moles and asking if it is correct is... funny. It is about as correct as writing

2=7

5=3.14

and so on.

Compare dilution and mixing calculations.
 
  • #5


Molarity = moles/liters so to revert to moles i would multiply by liters...
.2M x .002 liters = .0004 moles

Is this the problem?

Mr. Borek, where is it stated .2ml = .0004 moles?

Let's re establish the idea:

.002L of .2M "X" would by definition = .0004 moles "X" ... right?
The .0004 moles "X" are added to "Y" to create a .011L soln creating a .0004/.011 (moles/liter=molarity) = .0363636...M "X"... right?

Since I've been confused by the above responses could some one confirm or deny my results and suggest a formula to obtain the correct one if I am wrong?

Thanks,

Whalstib
 
Last edited:
  • #6


Whalstib said:
Molarity = moles/liters so to revert to moles i would multiply by liters...
.2M x .002 liters = .0004 moles

Is this the problem?

Borek was commenting on the way you set out your working, not the actual math and chemistry involved.

If your first line was "2.0ml=.002L", then it would be ok; but you instead had "2.0ml=.002L x .2M". 2mL is indeed equivalent to 0.002L, but it is not the same thing as 0.0004 moles. If you ever use an equality sign in your working, you should make sure that the quantities on either side have the same dimensions.

If i was doing this problem, i'd break it into multiple lines:
2mL = 0.002L
0.002L * 0.2 mol/L = 0.0004 moles of X
0.0004 moles / 0.011L = 0.03636 mol/L

Or something similar.

EDIT: I must have posted this just after you edited the post above.
 
  • #7


Whalstib said:
Mr. Borek, where is it stated .2ml = .0004 moles?

My fault with not quoting you exactly, but you started with

Whalstib said:
2.0ml=.002L x .2M

Which is equivalent of stating 2mL=0.0004 mole. That's an obvious shortcut, and notation is awkward - it suggests something you don't mean.

Whalstib said:
.002L of .2M "X" would by definition = .0004 moles "X" ... right?
The .0004 moles "X" are added to "Y" to create a .011L soln creating a .0004/.011 (moles/liter=molarity) = .0363636...M "X"... right?

0.036M is a correct result. And theory behind (which is not different from what you did) is on the page I linked to.

Note that volumes are not additive, so in general it may happen that 2mL + 9mL are not 11 mL. Unlikely in the case of diluted solutions, and rarely worth of taking into account, but it may hit one day if you ignore it.

--
methods
 
  • #8


Thanks!

Sorry for the sloppy notation.

I'll try to work on that as I have LOTS of questions this semester!

Whalstib
 

FAQ: Simplified 2.0 mL .20M X added to 9ml

What does "Simplified 2.0 mL .20M X added to 9ml" mean?

"Simplified 2.0 mL .20M X added to 9ml" is a statement that describes a chemical reaction. It means that 2.0 milliliters of a solution containing 0.20 moles of substance X is added to 9 milliliters of another solution in order to create a new mixture.

What is the purpose of adding "Simplified 2.0 mL .20M X" to 9ml?

The purpose of adding "Simplified 2.0 mL .20M X" to 9ml is to create a new mixture with a specific concentration of substance X. This mixture can then be used for various experiments or analysis.

How do I calculate the final concentration of substance X in the new mixture?

To calculate the final concentration of substance X in the new mixture, you can use the formula C1V1 = C2V2, where C1 and V1 are the initial concentration and volume of substance X, and C2 and V2 are the final concentration and volume of the new mixture. In this case, C1 = 0.20M, V1 = 2.0 mL, C2 = unknown, and V2 = 11 mL (9 mL + 2 mL). Solving for C2, we get a final concentration of 0.036M.

What precautions should I take when working with "Simplified 2.0 mL .20M X added to 9ml"?

When working with "Simplified 2.0 mL .20M X added to 9ml", it is important to wear appropriate protective gear such as gloves and goggles. You should also work in a well-ventilated area and avoid skin contact with the solutions. Additionally, make sure to accurately measure and mix the solutions to avoid any errors or accidents.

What other information should I know about "Simplified 2.0 mL .20M X added to 9ml"?

There are a few other important pieces of information to know about "Simplified 2.0 mL .20M X added to 9ml". First, it is important to know the properties and hazards of substance X in order to handle it safely. Additionally, the final volume of the new mixture will be 11 mL, so make sure to use a container that can hold at least this volume. Lastly, the reaction between the two solutions may produce heat or gas, so be prepared to handle any potential side effects.

Back
Top