Solution:
Let we have
Amount of glucose, n2 = 0.15 mol; Volume of solution, V = 1000 mL
Mass of glucose, m2 = n2M2 = (0.15 mol) (180 g mol -1) = 27.0 g.
Mass of solution, m = V = (1000 mL) (1.10 g mL-1 ) = 1100 g
Mass of water, m1 = m1 - m2 = 1100 g - 27 g = 1073 g
Amount of water,
Mole fraction of glucose,