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    Calculate the concentration in both mol dm-3 & g dm-3:

    1.05g of NaOH in 500cm3 of solution
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    gdm-3 = 1.05/(0.5)
    For moldm-3:
    n=m/mr to find number of moles. Then use c=n/v
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    (Original post by Lazybutmotivated)
    Calculate the concentration in both mol dm-3 & g dm-3:

    1.05g of NaOH in 500cm3 of solution
    (Original post by d010534)
    gdm-3 = 1.05/(0.5)
    For moldm-3:
    n=m/mr to find number of moles. Then use c=n/v
    This is correct but a quick way to do the last part is to recognise that the difference between the two is that it goes from g to mol, and using the equation n=m/Mr you could just divide the answer got for g dm-3 by the Mr of NaOH
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    (Original post by KaylaB)
    This is correct but a quick way to do the last part is to recognise that the difference between the two is that it goes from g to mol, and using the equation n=m/Mr you could just divide the answer got for g dm-3 by the Mr of NaOH
    Using your method I would get 0.0525 g dm-3 (2.1/40=0.0525) however on the markscheme it says the answer is 84
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    (Original post by Lazybutmotivated)
    Using your method I would get 0.0525 g dm-3 (2.1/40=0.0525) however on the markscheme it says the answer is 84
    You should get 2.1 g dm-3 (doing 1.05 g / 0.5 dm3)
    Then divide 2.1 g dm-3 by the Mr of NaOH which is 40 which should give you 0.0525 mol dm-3 as you correctly calculated.
    I do not see where the 84 could have come from. Perhaps could you post the mark scheme so that I could have a look?
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    (Original post by KaylaB)
    You should get 2.1 g dm-3 (doing 1.05 g / 0.5 dm3)
    Then divide 2.1 g dm-3 by the Mr of NaOH which is 40 which should give you 0.0525 mol dm-3 as you correctly calculated.
    I do not see where the 84 could have come from. Perhaps could you post the mark scheme so that I could have a look?
    I

    I THInk the mark scheme is wrong. These questions are from chemsheet & I haven noticed a few other times in which the marksheme is incorrect
 
 
 
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