# Homework Help: Calculating concentration in molLl^-1 of a solution

1. Sep 25, 2011

### leah3000

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
no of mols= Mass/Mr

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

3. 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?

Last edited: Sep 25, 2011
2. Sep 25, 2011

### Staff: Mentor

Looks OK.

3. Sep 26, 2011

### leah3000

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

4. Sep 27, 2011

### Staff: Mentor

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 lets you calculate number of moles of substance in given amount of solution of a given concentration.