% 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.
What is the mole fraction of NaBr in an aqueous solution that contains 20.1% NaBr (weight/weight%)?
3 answers
0.0422
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 :)
1 answer