a.)A solution is prepared by dissolving 10.50g of CaCl2 in 100 mL of water at 25

Question

a.)A solution is prepared by dissolving 10.50g of CaCl2 in 100 mL of water at 25.1 Celsius. What is the molality of the solution? b.)Assuming that all of the solute breaks up into ions, at what temperature would this CaCl2 solution freeze? c.) The freezing point of the solution was measured as -4.68 C. what is the experimental value of the Van't Hoff coefficient? d.) Sulfuric acid (H2SO4) dissociates to Sulfate and two protons (H+) in solution. The Van't Hoff coefficient for sulfuric acid has been measured as 2.28. Explain. e.) Another solution is prepared by dissolving 0.7500g of an unknown non-electrolite in 15g of Carbon tetra-chloride(CCl4, Kf=-30 C/molal). The freezing point of this solution is 6.50 C lower than the freezing point of pure carbon tetrachloride. What is the molar mass of the unknown?

Explanation / Answer

a.) molality is mol/kg MW or CaCl2 is 110.984g/mol so 10.50g CaCl2 / 110.984 g/mol =0.0946 mol per divided by .1kg = .946 m solution b.) if the solution breaks up the freezing point would be -14.8 C CaCl2, i = 3 because one mol CaCl2 releases 1 mol Ca2+ and 2 mol Cl- for a total of 3 mol of particles. dT = 3 x 1.86 C/molal x .946 molal = 5.28 C The freezing point of the solution: 0 - 5.28 C = -5.28 C c.)1.712 3*x*.946=4.68 solve for x

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