Solutions

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nelms6678
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

Solutions

Postby nelms6678 » Fri Dec 08, 2017 2:34 pm

Practice Question: SOLUTIONS
You need to make a stock solution of 0.0800M of sodium carbonate-Na2CO3
How much sodium carbonate would you add to a 250ml volumetric flask before filling to 250ml with water to make a 0.0800M aqueous solution of sodium carbonate?

I know you should multiply 0.0800M x 250ml which gives 0.02.
Why is the final step to multiply that value (0.02) by the Molecular weight (105.98)?

nelms6678
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

Re: Solutions

Postby nelms6678 » Fri Dec 08, 2017 2:40 pm

Practice Question (pt.2)

How many grams of NaOH do you need to prepare a 3.00 x 10^2 mL stock solution of 2.00M NaOH?

so first 2.00M x 3L which gives 0.6.
Why do you multiply this value by NaOH Molecular Weight (40.069)? Shouldn't you divide because of the formula MI VI = MF VF?

Courtney Cheney 3E
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Re: Solutions

Postby Courtney Cheney 3E » Fri Dec 08, 2017 2:45 pm

0.02 is the amount of moles of sodium carbonate that need to be added to the flask, however you then need to multiply the number of moles of sodium carbonate by the molecular weight of sodium carbonate so you know how many grams you need to add to the flask to make your stock solution.

nelms6678
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

Re: Solutions

Postby nelms6678 » Fri Dec 08, 2017 2:49 pm

Practice Question (pt.3)

How many milliliters of the stock solution would you need to make 250mL 0.650M NaOH solution?
Im not sure how to go about this problem? Any advice?

Courtney Cheney 3E
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Re: Solutions

Postby Courtney Cheney 3E » Fri Dec 08, 2017 2:54 pm

An easy way to know that you have to multiply the number of moles by the molecular weight is by looking at what the value is being expressed in. So for example: O.02 is expressed in moles, and molecular weight is expressed in grams / mole so to get moles to grams you have to multiply the number of moles by grams per mole: (_#_ mole)( _#_grams / mole) moles will cancel out and give you the answer in grams.

Courtney Cheney 3E
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Re: Solutions

Postby Courtney Cheney 3E » Fri Dec 08, 2017 3:14 pm

Part 3 reply:
So for part 3 you are using the stock solution of 2.00M NaOH from part two to make 250mL (250mL X 1000mL/ 1L = 0.250L) of a 0.650M NaOH.
For this question you need to use the (Mi)(Vi) = (Mf)(Vf)
Given:
Mf = 0.650 M
Vf = 0.250 L
Mi = 2.00 M
Vi = ???
So you plug the values in..... (2.00 M) (Vi) = (0.650 M) (0.250 L)
This can be rearranged as ..... (Vi) = (0.650 M) (0.250) / (2.00 M)

And then you solve so Vi= 0.0812 L or 81.2 mL
So you would need 81.2 mL of the 2.00 M solution of NaOH to make 250 mL of a 0.650 M solution of NaOH


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