Answer to Question #20156 in Physical Chemistry for rahul
calculae the freezing point of a solution ,contaning 18 gram glucose and 68.4 gram sucrose in 200 gram of water.freezing point of pure water is 273 kelvin and cryoscopic constant for water is 1.86 K/m
Use the formula for freezing point depression.
Change in T = -Kf x molality of solution
In this solution, molality (b) = (18 /180)/0.2 L + (68.4/342 )/0.200 L = 0.3 mol per 0.2 L = 1.5 moles/1 kg water = 1.8 moles sucrose / kg water.
Change in T = - 1.86 C/m * 1.5 moles/1 kg water = -2.79 C