Asked by Mike
A solution contains 40.0 g of sodium chloride and 200.0 g of water. Calculate the NaCl
concentration as a a) mass percentage, b) molality, c) mole fraction.
I have trouble finding mole fraction. I believe I've got the other two, but I'm not 100% sure.
concentration as a a) mass percentage, b) molality, c) mole fraction.
I have trouble finding mole fraction. I believe I've got the other two, but I'm not 100% sure.
Answers
Answered by
DrBob222
% w/w = 16.7
m = 3.42
If you didn't get these show your work and I'll find the error.
mols NaCl = grams/molar mass
mols H2O = grams/molar mass
XNaCl = mols NaCl/total mols.
I worked it quickly and obtained 0.381 but you should confirm that.
As a check run XH2O and see if you get 1-0.381 because XNaCl + XH2O must = 1.00(within rounding error).
m = 3.42
If you didn't get these show your work and I'll find the error.
mols NaCl = grams/molar mass
mols H2O = grams/molar mass
XNaCl = mols NaCl/total mols.
I worked it quickly and obtained 0.381 but you should confirm that.
As a check run XH2O and see if you get 1-0.381 because XNaCl + XH2O must = 1.00(within rounding error).
Answered by
Mike
Dr.Bob, Could you please show the work for w/w and mol frac. I continue to get a completely bizarre number. The only one that I am getting correct is the molal part.
Answered by
Mike
I was able to find both the Mol Frac and m now. The only problem is w/w.
m) (40.0gNaCl)*(1molNaCl/58.33gNaCl)=.684mol Nacl
-200.0gH20*(1molH20/18.02gH20)=11.mol H20
-200g= .2L
** .684mol NaCl/.2LH20= m=3.42
Mol Frac) .864molNaCl/(.864molNaCl+11.02)= .580
w/w .684/11.02= 3.63
m) (40.0gNaCl)*(1molNaCl/58.33gNaCl)=.684mol Nacl
-200.0gH20*(1molH20/18.02gH20)=11.mol H20
-200g= .2L
** .684mol NaCl/.2LH20= m=3.42
Mol Frac) .864molNaCl/(.864molNaCl+11.02)= .580
w/w .684/11.02= 3.63
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