250 g solution of D -glucose in water contains 10 . 8 % of carbon by weight. The molality of the solution is nearest to (Given: Atomic Weights are H = 1 u ; C = 12 u ; O = 16 u )

Step-by-step solution

C 6 H 12 O 6 → Glucose Given: % C = 10 . 8 = mass   of   C mass   of   solution × 100 10 . 8 × 250 100 = mass of C ⇒ Mass of C = 27 gm mass   of   C mass   of   glucose = 72 180 mass   of   glucose   =   27 × 180 72 ∴ mass of glucose = 67 . 5 gm ∴ moles of glucose = 67 . 5 180   =   0 . 375 moles Mass of solvent = 250 - 67 . 5 gm = 182 . 5 gm ∴ Molality = moles   of   solute mass   of &#