Question

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!

Answer 1

The wording is a bit ambiguous, to me.

Normally, one would want the mole fraction of the <u> solutes</u>, 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

Answer 2

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

Answer 3

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

Answer 4

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?

Answer 5

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

Answer 6

Nuts to that.

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

Answer 7

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