Question

The mole fraction of glucose (C6H1206) in an
aqueous binary solution is 0.1. The mass
percentage of water in it, to the nearest
integer, is ....​

Answer 1

Answer:

Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9 so mass % of water = (0.9 x 18)/(0.1 x 180 + 0.9 x 18) x 100 = 47.36 = 47