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AP chemistry

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8.3 grams of NaCl and 3 grams of KBr were dissolved in 47 mL of water. What is the mole fraction of KBr in the solution?

I know that the mole fraction of the KBr and Nacl (without using the mL H2O is .2, but, i am confused on where i use that mL in the problem.) Please help!

  • AP chemistry - ,

    The wording is a bit ambiguous, to me.

    Normally, one would want the mole fraction of the solutes, which would be

    fraction=molesKBr/(molesKBr+molesNaCl)

    However, the wording is not as specific as this. As it is worded, one has to inclue the moles of water, so
    fraction=MolesKBr/(moleswater+molesNaCl+molesKBr)


    How do you get moles of water?
    moleswater=densitywater * volumewater/molmass water

  • AP chemistry - ,

    i got 1.2 but i don't think i did it right. below is my work:
    3g(1mole/199g)=.03moles KBr
    .03/(1moleH2O+.03molesKBr+.14moles NaCl)= 1.2

    moles H2O = (1(47))/47=1

  • AP chemistry - ,

    nevermind,i got .02...did i do this correctly??

  • AP chemistry - ,

    Your math is right, however, in the 3 grams (one significant figure), I suspect the writer was sloppy, and meant at least 3.00 grams
    3.00/199=.00151moles. I don't know how you got .03 out of that.

    How did you get moles water as 1?

  • AP chemistry - ,

    d=m/v=47g/47mL
    because 1g water = 1mL water
    moles= dv/m=1(47)/47=1

  • AP chemistry - ,

    Nuts to that.

    Moles water= masswater/molmasswater=47g/18g/mol= not 1

  • AP chemistry - ,

    so, i got 6.82 moles of water and plug that in and get .00361, right?

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