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    any ideas? In the question i need to work out what the carbonate is. I understand how to work out the moles but why do i need to times the moles by 10 to get it to the moles in 250cm3 solution?

    Solution G - HCl - 0.150 mol dm-3

    Solution H - X2CO3 - 75.95 g dm-3

    X2CO3(aq) + 2HCl(aq) ---> 2XCl(aq) + CO2(g) + H20(l)

    Titre value of G: 18.5 cm3

    Thank you in advance
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    (Original post by Appazap)
    any ideas? In the question i need to work out what the carbonate is. I understand how to work out the moles but why do i need to times the moles by 10 to get it to the moles in 250cm3 solution?

    Solution G - HCl - 0.150 mol dm-3

    Solution H - X2CO3 - 75.95 g dm-3

    X2CO3(aq) + 2HCl(aq) ---> 2XCl(aq) + CO2(g) + H20(l)

    Titre value of G: 18.5 cm3

    Thank you in advance

    Did you pipette 25cm3 of solution H? Was there originally 250cm3 of solution H?
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    (Original post by TutorsChemistry)
    Did you pipette 25cm3 of solution H? Was there originally 250cm3 of solution H?
    yeah that was the standard solution i used.
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    Thought so.

    Your calculation will work out how many moles are in the 25cm3 (because that's the quantity you titrated).
    You need to know how many moles were in the entire sample, in this case in the 250cm3. So we need to scale up to the size of the entire sample. In this example the factor to scale up is 10 (250/25).
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    (Original post by TutorsChemistry)
    Thought so.

    Your calculation will work out how many moles are in the 25cm3 (because that's the quantity you titrated).
    You need to know how many moles were in the entire sample, in this case in the 250cm3. So we need to scale up to the size of the entire sample. In this example the factor to scale up is 10 (250/25).
    Ooooooooh thank you so much
 
 
 
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