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Pleez Help Me

  1. Nov 10, 2005 #1
    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!!
  2. jcsd
  3. Nov 10, 2005 #2


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    How would you relate 25cm3 to moles using molarity?
  4. Nov 10, 2005 #3

    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??
  5. Nov 10, 2005 #4


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    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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