Calculate the molarity of the sodium carbonate solution

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Discussion Overview

The discussion revolves around calculating the molarity of a sodium carbonate solution based on a given mass and volume. Participants are exploring the steps involved in solving this problem, including the conversion of units and the relationship between moles and molarity.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant requests assistance in tackling a problem involving the molarity of a sodium carbonate solution, expressing a desire to understand the steps for similar questions.
  • Another participant asks how to relate the volume of 25 cm3 to moles using molarity, indicating a need for clarification on the concept.
  • A similar question is posed again regarding the relationship between volume and moles, suggesting a focus on understanding the formula for molarity.
  • One participant suggests calculating the number of moles of sodium carbonate by dividing the mass (5.3 g) by its molar mass, and explains how to derive molarity from moles and volume.
  • The explanation includes the conversion of volume from cm3 to liters and the formula for calculating molarity as moles of solute divided by liters of solution.
  • There is an emphasis on working backward from known concentration to find moles, reiterating the relationship between moles, volume, and molarity.

Areas of Agreement / Disagreement

Participants are generally aligned on the approach to calculating molarity, but there is no consensus on the specific steps or methods to be used, as some questions remain unanswered and require further clarification.

Contextual Notes

Some participants express uncertainty about the conversion of units and the application of the molarity formula, indicating that additional assumptions or definitions may be necessary for a complete understanding.

suf7
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Pleez Help Me!

when a solution of sulphuric acid was standardised the following results were obtained. exactly 5.3g of pure anhydrous sodium carbonate were dissolved in wter and made up to 250cm3 of solution. 25cm3 of this solution were neutralised by 20cm3 of the acid. calculate the molarity of the sodium carbonate solution.

can sum1 show me how2tackle this question pleez??..i wana know what are the steps involved so i can do similar questions on my own..what do i do with this??..thank u!
 
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How would you relate 25cm3 to moles using molarity?
 
GCT said:
How would you relate 25cm3 to moles using molarity?


is molarity = moles over volume??
so 25cm3 needs to be converted into dm3 by dividing by 1000 then how wud i work out the moles??
 
You know the mass of Sodium Carbonate (5.3 g). You can look up/calculate the molar mass. Divide the amount of Na2CO3 you have by the molar mass to get the number of moles you have.

5.3 grams of Na2CO3 is dissolved in 250 cm^ of solution.
Molarity is moles of solute (in this case Na2CO3) / Liters of solution.
1 liter = 1000 cm^3

to get the number of moles of Na2CO3 there are in a certain volume of solution with a known concentration, remember how to calculate the concentration and work backward.
moles / volume = molarity, so molarity * volume = moles
 

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