Question

Part B A solution of water (kr = 1.86°C/m) and glucose freezes at - 415°C What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 °C Express your answer to three significant figures and include the appropriate units. View Available Hint(s) m= 2.34 m Submit Previous Answers X Incorrect: Try Again: 5 attempts remaining

Answer 1

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Part B. KF = 1.86.0lm freezing point of pure water = TE = 0.00°C freezing point of solution Tf = -4.15°C depression, freezing point = ST = TEO - TE AT = 0.00'C - (-4150) = 4.15 °C is van't Hoff factor i= 1 for glucose m= molal concentration of glucose in the solution We know that AT = iKfm 415C = 1 X 1.86C)m X m m= 2.23.m m= 2.23m Answer Part c. Kb = 0.512.01m boiling point of bure water & Too = 100.00.0 boiling point of solution To= 101.56°C o elevation in boiling point = AT = To - Too AT = 101.560 - 100.000 AT= 1.56.c I= van't Hoff factor i = 1 for glucose m= molal concentration of glucose in the solution We know that UT = ikom 1.56.C = 1 x 0.512.01m x m m = 3,04 m