Let mass of solution = 100 g ∴ Mass of glucose = 10 g Mass of water = 90 g = 0.09 kg No. of moles in 10 g glucose $= \frac{10}{180}$ = 0.0555 mol No. of moles in 90g $\mathrm{H_2O}$ $= \frac{90}{18}$ = 5 molesVolume of solution $= \frac{100\,\text{g}}{1.2\,\text{gmL}^{-1}}$ = 83.33 mL = 0.0833 L(a) Molality $= \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$ $= \frac{0.0555\,\text{mol}}{0.09\,\text{kg}}$ = 0.617 m(b) x (Glucose) $= \frac{\text{Number of moles of solute}}{\text{Number of moles of solution}}$ $= \frac{0.0555}{5.0555}$ = 0.01∴ x $(\mathrm{H_2O})$ = 1 – 0.01 = 0.99(c) Molarity $= \frac{\text{Number of moles of solute}}{\text{Volume of solution in L}}$ $= \frac{0.0555}{0.0833}$ = 0.666 M