Question

When a certain liquid of molar mass 46.1 g/mol freezes at -3.65 degrees celsius its density changes from 0.789 g/cm^3 to 0.801 g/cm^3. Its enthalpy of fusionis 8.68 kJ/mol. Estimate the freezing point of the liquid at 100 MPa. 

Answer 1

29 gem - 3 # Density of the liquid , I, = 0.789 gem- 20.7899 cms x 103 си -3 = 789 kg m-3 Density of the solid, s 20.30lgam-3. = 801 kg m- img3 Molan mass of liquid, M = 46.1 gmail- & 1kg 10g od 20.0461 kg mol- The liquid freezes at -3.65°c. . Tz -3.65°C bila = 1-3.65+273.15)X = 269.5k.. Enthalpy of fusion, Altus = 8.68 kJ mol- Enthalpy of freezing is heus negative of Althus isho = - 1143 -8.68 kJ medel. . change in molar volume Do = * M ( 1 2 3 4 +7(0.04 kymo ) (Bolg mit is my me) = -8.75 710-7 m3 marts

The initial pressing at which free ring takes place a laten 7 pe latm - 1 atm x 101.3 kPa - lalm. 2101.3 x pa let Freening point of liquid at 100 MPa is to This pressure p = 100 MPa. 100MPa x 16² kPa - 1 MPa. 2105 K Pa. Now, we know it Thomson - James equation is - .. plz p A AT A 17 Dhars mm on, 1'249.5 x + [lvou a -10.3 x A). 219.52.1-8.95% -8.68 kJ mal? 2 269.5 K + 2.71K. 2272.2. The Auswen - The freezing point of liquid at 100 MPa is 272.2K.