A 5 % solution (by mass) of cane sugar in water has freezing point of 271 K . Calculate the freezing point of 5 % glucose in water if freezing point of pure water is 273 . 15 K .

Sugar cane contains sucrose thus molar mass of sugar = 342 g mol - 1 The 5 % of sugarcane solution means 5 g of sucrose in 100 g of solution. Weight of solvent can be calculated as follows: ∴ w 1 = 100 - 5 = 95 g Molality of sugar solution = w 2 / M 2 w 1 / 1000 = 5 × 1000 342 × 95 = 0 . 1538 m ∆ T f for sugar solution = 273 . 15 - 271 = 2 . 15 K ∆ T f = K f × m K f = 2 . 15 0 . 1538 = 13 . 9791 Km - 1 Molar mass of glucose is 180 g / mol The 5 % of glucose solution means 5 g of glucose in 100 g of solution. Weight of solvent can be calculated as follows. ∴ w 1 = 100 - 5 = 95 g molality = w 2 / M 2 w 1 / 1000 = 5 180 × 1000 95 = 0 . 2924 m ∆ T f = K f × m = 2 . 15 0 . 1538 × 0 . 2924 = 4 . 09 K Freezing point of glucose = 273 . 15 - 4 . 09 = 269 . 06 K

An antifreeze solution is prepared from 222 . 6 g of ethylene glycol C 2 H 6 O 2 and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1 . 072 g mL - 1 , then what shall be the molarity of the solution?

Mass of solute(Ethylene glycol) = 222 . 6 g The molar mass of solute (Ethylene glycol) C 2 H 4 OH 2 = 12 × 2 + 4 + 2 16 + 1 = 62 g mol - 1 Number of moles of solute= Mass of solute Molar mass = 222 . 6 62 = 3 . 59 Mass of solvent= 200 g Molality = Moles of solute Mass of solvent in grams × 1000 = 3 . 59 200 × 1000 = 17 . 95 mol kg - 1 Total mass of the solution = 422 . 6 g Volume of the solution= Mass of the solution density of the solution = 422 . 6 1 . 072 = 394 . 21 ml Molarity= Number of moles of solute Volume of solution × 1000 = 3 . 59 394 . 2 × 1000 = 9 . 1 mol . L - 1