Calculating concentration in molLl^-1 of a solution

[leah3000]

Homework Statement

Determine the concentration in molL^-1 of a solution of sodium carbonate whoch contains 53g of sodium carbonate in 500ml of solution. Calculate the concentration of sodium and carbonate ions in this solution.

Homework Equations

no of mols= Mass/Mr

mols= [ ] x vol
----------
1000

The Attempt at a Solution

Mr= 106
no of mols= 53/106 = 0.5mols

molsx 1000/ vol = [ ]

0.5x1000/500= 1molL^-1

Na2CO3

2mols Na^+: 1mol CO3^2-

I'm usure of how to calculate the rest.

would the conc of Na ion be 2x 1molL^-1 = 2molL^-1

then making the conc of CO3^2- just 1molL^-1?

 

[Borek]

Looks OK.

 

[leah3000]

Looks OK.

thanks you...

There's another question based on the answer to the first but I'm not sure i understand what it says

How would prepare a 250ml of a .25M solution of sodium carbonate using the sodium carbonate in the quest above. show claclulations.

This is what I did.

Mass= Molsx Mr

Mols= 0.25x1000/250

=1mol

Mass= 1x106
=106g

0.25M= 106g Na2CO3^2- in 1000ml
In 250ml =106/1000 x250 =26.5g

Therefore to make up the 250ml of .25M Na2CO3^2- add 26.5g Na2CO3^2- to 250 ml

 

[Borek]

Mols= 0.25x1000/250

=1mol

Not sure what you are doing here, 250 mL of 0.25 M solution doesn't contain 1 mole of substance.

By definition molar concentration is

C = \frac n V

solve for n to find the formula that let's you calculate number of moles of substance in given amount of solution of a given concentration.