25 grams of sodium floride is fixed with 200 grams of water . what is the mole fraction of sodium floride in the solution.
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Answer:
0.0468
Explanation:
no.of mol of NaF = 25/42= 0.595
no.of mol of water = 200/18 = 11.11
mol fraction of NaF in solution =
0.595/(0.595+11.11)
= 0.595/12.705
= 0.0468
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