G.5  [ENDORSED]

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Fiona Grant 1I
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Joined: Fri Apr 06, 2018 11:03 am

G.5

Postby Fiona Grant 1I » Sun Apr 08, 2018 1:33 pm

I am still a bit confused as to the setup of this problem. How do you find the molarity of the Na2CO3 solution using the mass and volume given? I know that n=m/M and c=n/V, but I don't understand how to find the concentration using these equations.

Natalie Noble 1G
Posts: 30
Joined: Thu Feb 01, 2018 3:02 am
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Re: G.5  [ENDORSED]

Postby Natalie Noble 1G » Sun Apr 08, 2018 1:35 pm

Molarity = moles/volume

You are given 2.111g Na2Co3 but to put it in the formula above you need to find how many moles there are of Na2Co3

N = m/M or
Moles = mass/molar mass

105.99g/mol is the molar mass of Na2Co3

Moles = (2.111g) / ( 105.99g/mol)
Moles = .0199 mol

Now you have the moles of Na2Co3, and the problem gave you the volume of the solution (250ml) but since molarity is mol/L convert 250ml in L, which becomes .250L

Molarity = moles/volume

Molarity = .0199 mol / .250L
Molarity = .07967M


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