278k时甘油水溶液的渗透压是1215.6kPa,计算该溶液的凝固点

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由范特荷甫方程
πV = nRT,c = n/V
解得物质的量浓度c = π/RT = 1215600 ÷(8.314×278) = 525.94 mol/m^3
溶液密度ρ ≈ 1 g/cm^3 = 10^3 kg/m^3
所以溶液的质量摩尔浓度 m= c/ρ 525.94×10^-3 mol/kg
Kf(水) = 1.86 K·kg/mol
凝固点的下降△Tf = Kf × m = 1.86 K·kg/mol × 525.94×10^-3 mol/kg = 0.98 K
所以凝固点是273.15 - 0.98 =272.17 K
======以下答案可供参考======
供参考答案1:
基本思路：根据渗透压求得浓度，再根据浓度求出其凝固点(T')
Pi = cRT
273 - T' = Kf*c = Kf*Pi/(RT)
T' = 273 - Kf*Pi/(RT)
将Kf = 1.86, Pi = 1215.6kPa, R=8.314, T = 278K代入得：
T' = 273 - 1.86*1215.6/(8.314*278) = 272 K