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Math Help - basic chemistry formulas

  1. #1
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    basic chemistry formulas

    Having problems figurring out my math part of chemistry. Please someone help me.
    stressed.

    How many moles of NaCl are dissolved in 395 mL of a 1.20M NaCl solution? i am not sure how i am to write this problem out.
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  2. #2
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    Smile

    hi
    In general we have : n(moles) = c\times v
    here you have C = 1.20 (mol/l) and V = 0.395 (l)
    you do the math
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  3. #3
    MHF Contributor alexmahone's Avatar
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    1.20 M = 1.20 mol/litre
    395 ml=395*10^{-3} litre

    Number of moles= 1.20*395*10^{-3}=0.474
    Last edited by alexmahone; June 30th 2009 at 12:30 AM.
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  4. #4
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    omg

    you have to be kidding. here i was trying so hard and it was just that easy.

    you are the bomb. thank you
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  5. #5
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    i have several questions

    how many moles of Al(NO3)3 would be contained in a 15.5 cc sample of a 0.250 M Al(NO3)3 solution> ? ok now will i need to get the atom mass of the element to use in this problem? will i use the mole 6.022 x1023?

    how will it write it. i am online chemistr but lab 6 hrs on monday. only see teacher for those 6 hrsx. but busi with lab and no time for problem solving
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  6. #6
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    Quote Originally Posted by troubledmind View Post
    how many moles of Al(NO3)3 would be contained in a 15.5 cc sample of a 0.250 M Al(NO3)3 solution> ? ok now will i need to get the atom mass of the element to use in this problem? will i use the mole 6.022 x1023?

    how will it write it. i am online chemistr but lab 6 hrs on monday. only see teacher for those 6 hrsx. but busi with lab and no time for problem solving
    You can use latex to represent chemical formulae here: Al(NO_3)_3. If you have any more chemistry questions there is also CHF (there is a link in my sig)

    For this question you do not need to use Avogadro's number because it's asking for a value of moles rather than molecules. You do not need the atomic mass of Al(NO_3)_3 either since the question asks for the moles and the formula used give you a molar answer.

    To solve this you may either use the method outlined in post 2 or you can use the formula:

    n = \frac{MV}{1000}

    Where
    • n is the number of moles
    • M is the molarity in mol \: dm^{-3}
    • V is the volume in cm^3


    Plugging in the numbers gives: n = \frac{0.250 \times 15.5}{1000} = 3.875\times 10^{-3} \text {{ mol }}
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  7. #7
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    Quote Originally Posted by troubledmind View Post
    Having problems figurring out my math part of chemistry. Please someone help me.
    stressed.

    How many moles of NaCl are dissolved in 395 mL of a 1.20M NaCl solution? i am not sure how i am to write this problem out.
    Post chemistry questions here: http://www.chemistryhelpforum.com/chemistry-help/
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