Vitamin K is involved in normal blood clotting. When 2.19 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point of the solution is lowered by 7.33 °C. Look up the freezing point and Kf constant for camphor in the Colligative Constants table. Calculate the molar mass of vitamin K. molar mass: g/mol Colligative Constants Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Formula K, value* Normal freezing K, value Normal boiling point (°C) Solvent (°C/m) (°C/m) point (°C) water H20 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol -117.3 C2H60 CCIA 1.99 1.22 78.4 carbon 29.8 -22.9 5.03 76.8 tetrachloride camphor C10H160 37.8 176

Vitamin K is involved in normal blood clotting. When 2.19 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point of the solution is lowered by 7.33 °C. Look up the freezing point and Kf constant for camphor in the Colligative Constants table. Calculate the molar mass of vitamin K. molar mass: g/mol Colligative Constants Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Formula K, value* Normal freezing K, value Normal boiling point (°C) Solvent (°C/m) (°C/m) point (°C) water H20 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol -117.3 C2H60 CCIA 1.99 1.22 78.4 carbon 29.8 -22.9 5.03 76.8 tetrachloride camphor C10H160 37.8 176

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Quinine is a natural product extracted from the bark of the cinchona tree, which is native to South America. Quinine is used as an antimalarial agent. When 1.90 g of quinine is dissolved in 25.0 g of cyclohexane, the freezing point of the solution is lowered by 4.88 °C. Look up the freezing point and Kf constant for cyclohexane in the Colligative Constants table. Calculate the molar mass of quinine. molar mass: g/mol
The boiling point of an aqueous solution is 101.45 Celsius. What is the freezing point?
Assuming 100% dissociation, calculate the freezing point and the boiling point of 0.570 mol K3PO4 in 1.00 kg water. You may need the colligative constants for water.
In the temperature-composition diagram shown here, the vaporization temperature of the pure solute is T3. 00 True False トピド T To T₁ T2 T3 N Ο Χ1 X2 X3 X4 1 Xsolute
Calculate boiling point depression and freezing point depression of .25 m K2SO4.
Colligative Constants Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Kvalue Normal freezing (°C/ m) point (°C) water H₂0 1.86 benzene C6H6 5.12 cyclohexane C6H 12 20.8 ethanol C₂H6O 1.99 carbon tetrachloride 29.8 CCI 4 camphor C 10 H 16 0 37.8 Solvent Formula- 0.00 5.49 6.59 -117.3 -22,9 176 K, value Normal boiling (°C/ m) point (°C) 0.512 100.00 2.53 2.92 1.22 5.03 80.1 80.7 784 76.8
A solution is prepared using 19 mL of hexane and 23 mL of heptane. What is the mole fraction of hexane in the vapor above the solution? Provide your answer to the nearest hundredth. The information below may be useful to you: MW (g/moL) density (g/mL) Standard Vapor Pressure (kPa) hexane 86.15 0.655 20.1 heptane 100.21 0.684 6.1
Q5. The depression of the freezing point with FeCl3 s 0.2976 degC and molality is 0.050 m. Calculate the van't Hoff factor i for iron chloride, if the molal freezing depression point constant for water is 1.86 degC/m. Q6.
True or False1.) The phase diagram of a pure solvent shows a lower freezing point compared to solutions2.) The process of freezing only lowers the kinetic energy of solute particles in a solution3.) Ionic solutes have higher van't hoff factor compared to non ionic solutes4.) The formation of intermolecular forces of attraction requires complete dissolution
Referring to Table 13.3, calculate the freezing point of a solutioncontaining 0.600 kg of CHCl3 and 42.0 g of eucalyptol(C10H18O), a fragrant substance found in the leaves of eucalyptustrees.
A solution is prepared by dissolving 8.215 g of an unknown pure nonvolatile compound in 300.0 g of benzene (C6H6, MW = 78.11 g/mol). The freezing point of the solution is 0.38 °C lower than that of pure benzene. Based on this information, find the molecular weight of the unknown compound. Note that for benzene Tf° = 5.50 °C and Kf = 5.12 kg°C/mol.
A certain substance X has a normal freezing point of 4.1 °C and a molal freezing point depression constant Kf= 2.09 °C-kg-mol¯!. Calculate the freezing point of a solution made of 34.1 g of urea ((NH2), cO) dissolved in 550. g of X. Be sure your answer has the correct number of significant digits.