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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


    However, the wording is not as specific as this. As it is worded, one has to inclue the moles of water, so

    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 -

    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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