Question #20156

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

Expert's answer

Use the formula for freezing point depression.

Change in T = -K_{f} 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.

Therefore,

Change in T = - 1.86 C/m * 1.5 moles/1 kg water = -2.79 C

Change in T = -K

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.

Therefore,

Change in T = - 1.86 C/m * 1.5 moles/1 kg water = -2.79 C

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