Asked by Steve
What is the mole fraction of NaBr in an aqueous solution that contains 20.1% NaBr (weight/weight%)?
Answers
Answered by
DrBob222
% w/w of 20.1% means 20.1 g NaBr in 100 g solution.
So g NaBr = 20.1
g solution = 100 g = g NaBr + g H2O or
g H2O = 100-20.1 = 79.9
n NaBr = grams/molar mass = ?
n H2O = grams/molar mass = ?
Total mols = n NaBr + n H2O
XNaBr = nNaBr/total mols.
So g NaBr = 20.1
g solution = 100 g = g NaBr + g H2O or
g H2O = 100-20.1 = 79.9
n NaBr = grams/molar mass = ?
n H2O = grams/molar mass = ?
Total mols = n NaBr + n H2O
XNaBr = nNaBr/total mols.
Answered by
Ilizabeth
0.0422
Answered by
Joseph
No one has posted the solution and correct answer so for anyone who looks this up in the future.
Assume the solution is 100g
100-20.1=79.9
20.1g of NaBr and 79.9g of H2O
the atomic mass of NaBr is 102.89 and the atomic mass of H2O is 18.016
20.1/102.89=0.1954 79.9/18.016=4.4349
to get the total weight of the solution we add NaBr to H2O
0.1954+4.4349=4.6303
now we find the weight %
0.1954/4.6303=0.0422
hope this helps any future people :)
Assume the solution is 100g
100-20.1=79.9
20.1g of NaBr and 79.9g of H2O
the atomic mass of NaBr is 102.89 and the atomic mass of H2O is 18.016
20.1/102.89=0.1954 79.9/18.016=4.4349
to get the total weight of the solution we add NaBr to H2O
0.1954+4.4349=4.6303
now we find the weight %
0.1954/4.6303=0.0422
hope this helps any future people :)
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