G.5  [ENDORSED]

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Jade Corpus-Sapida 1G
Posts: 35
Joined: Fri Apr 06, 2018 11:02 am

G.5

Postby Jade Corpus-Sapida 1G » Fri Apr 06, 2018 11:15 am

Can someone explain why in part a of G.5 theres the multiplying of Na2CO3 mol and dividing of Na moles such as:
v= (2.15E-3 mol Na)(1 mol Na2CO3)/(0.0797 mol/L)(2 mol Na) ?
I'm confused as I know the formula is Volume=mol/conc and I see that it's asking in relation to Na. But then theres the compound's moles as well as the dividing of sodium moles. Is this just to relate the reactants to products?

Alma Carrera 3C
Posts: 33
Joined: Fri Apr 06, 2018 11:02 am

Re: G.5

Postby Alma Carrera 3C » Fri Apr 06, 2018 11:42 am

They are just multiplying by the ratio because for every mole of Na2CO2 there are 2 moles of Na.

Madeleine Farrington 1B
Posts: 32
Joined: Fri Apr 06, 2018 11:02 am

Re: G.5

Postby Madeleine Farrington 1B » Fri Apr 06, 2018 12:33 pm

I also had trouble with part C of this problem.

G.5.c: A student prepared a solution of sodium carbonate by adding 2.111g of the solid to a 250.0 mL volumetric flask and adding water to the mark. Some of this solution was transferred to a buret. What volume of solution should the student transfer into a flask to obtain 50.0 mg Na2CO3?

In obtaining this, the answer manual shows the process (50.0x10^-3 g Na2CO3)/(105.99g.mol^-1 x 0.0797 mol.L^-1 Na2CO3). Why do you divide by 105.99 g.mol^-1 and where is this value coming from?

Anna De Schutter - 1A
Posts: 66
Joined: Wed Feb 21, 2018 3:01 am

Re: G.5  [ENDORSED]

Postby Anna De Schutter - 1A » Fri Apr 06, 2018 1:21 pm

Hi Madeleine!

To come back to your question about G5 part c I think this is one way to look at it:

According to the key concept: When diluting a solution, moles of solute remain the same, we can write:

n(initial)=n(final)
or
M(initial)*V(initial)=M(final)*V(final)

In part c of problem G5 we want to solve for V(initial)

To start we know what M(initial) is:
M(initial)=n(initial)/V(initial).
n(initial)=mass(initial)/molar mass of Na2CO3

The molar mass of Na2CO3= (2*22.99g/mol)+(1*12.01g/mol)+(3*16.00g/mol)= 105.99g/mol (this is where that value is coming from).

As a result M(initial)=n(initial)/V(initial)=(mass(initial)/molar mass of Na2CO3)/V(initial)= ((2.111g)/(105.99g/mol))/(0.250L)= 0.07967mol/L

Normally we would then solve for V(initial) by doing V(initial)=(M(final)*V(final))/M(initial)

However, in this problem they approach it slightly differently.

We know:
M(initial)*V(initial)=M(final)*V(final)
But this also equal:
M(initial)*V(initial)=n(final) as n(final)=M(final)*V(final)

Thus V(initial)=n(final)/M(initial) We can solve this equation because we know what n(final) and M(initial) are.

Indeed M(initial)=0.07967mol/L

Now we just need to calculate n(final). We do this the same way as we did with n(initial).
n(final)=mass(final)/molar mass of Na2CO3=(50.0*10-3g)/(105.99g/mol).

Thus V(initial)=n(final)/M(initial)=(mass(final)/molar mass of Na2CO)/M(initial)= (50.0*10-3g)/(105.99g/mol)/(0.07967mol/L) which is the equation given in the solutions manual.

I hope this helps! :)
Anna De Schutter - section 1A


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