Calculating concentration in molLl^-1 of a solution

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Discussion Overview

The discussion revolves around calculating the concentration in molL-1 of a sodium carbonate solution, including the determination of concentrations of sodium and carbonate ions. It also addresses a follow-up question regarding the preparation of a specific molarity solution based on the initial calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • The initial calculation of the number of moles of sodium carbonate is presented as 0.5 moles, leading to a concentration of 1 molL-1.
  • Some participants suggest that the concentration of sodium ions would be 2 molL-1 based on the dissociation of sodium carbonate, while the concentration of carbonate ions would be 1 molL-1.
  • A follow-up question involves preparing a 250 mL solution of 0.25 M sodium carbonate, with calculations provided that suggest a mass of 26.5 g is needed.
  • One participant challenges the calculation of moles for the 0.25 M solution, indicating that the calculation presented does not align with the definition of molar concentration.

Areas of Agreement / Disagreement

There is no clear consensus on the calculations for the preparation of the 0.25 M solution, as one participant questions the accuracy of the mole calculation. The discussion includes differing interpretations of the results and calculations presented.

Contextual Notes

The calculations depend on the correct application of molarity definitions and may involve assumptions about the purity of the sodium carbonate used. The discussion does not resolve the discrepancies in the calculations presented.

Who May Find This Useful

Students and individuals interested in chemistry, particularly those learning about solution concentrations and molarity calculations.

leah3000
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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?
 
Last edited:
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Looks OK.
 
Borek said:
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
 
leah3000 said:
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.
 

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