When a solution of 1.00 g of ‘X’ in 100 ml of water was shaken with 10 ml of diethyl ether, 0.60 g of X was transferred to the solvent (ether) layer. When another solution of 0.5 g of ‘X’ in 100 ml of water was shaken with 5 ml of petroleum ether, 0.3 g of X was extracted by the solvent. Which of these ethers is a better solvent? Justify your answer using calculations.

My imaginary molecular compound has an empirical formula of C7H40. If you dissolve 2.64 grams of this compound in 22.8 g of a solvent with a K=20C/m and the freezing point is lowered by 11.15C, what is the molecular formula of this imaginary compound? Upload your work to the free response questions into the moodle assignment. Please upload one document for all of the questions and label this question Free Response-1. O CH4O O C14H8O2 O C21H1203 O C28H1604

1. Calculate the freezing point of a solution of 5.00 g of diphenyl, C12H10, and 7.50 g of naphthalene, C1oHs, dissolved in 200.0 g of benzene (Freezing Point = 5.50 °C; kfp = 5.12 °C /m).(NOTE: There is a tiny twist in this problem, but keep in mind that colligative properties are all about how many particles in solution (a.k.a., moles ©) and nothing else.

Given the following information, determine the percentage by mass of cyclohexane in the mixture. volume of the solution in mL 3.15 Vapor temperature when distillation of cyclohexane started (°C) 80.95 Vapor temperature when distillation of cyclohexane finished (°C) 83.09 Volume of cyclohexane collected in mL 1.60 Vapor temperature when distillation of toluene started (°C) 110.81 Vapor temperature when distillation of toluene finished (°C) 111.88 Volume of toluene collected in mL 1.55

In regards to freezing point depression:  The original experiment calls for 3 grams lauric acid and .4 g benzoic acid to be melted and then observed to crystalize to demonstrate freezing point depression.  A student misreads the directions and adds 4.0 g of benzoic acid instead of 0.4 g. What do you expect to happen? Would the experiment still be valid? Explain your reasoning. I think that since the amount of lauric acid (3g) doesnt change that the solution may be too oversaturated with benzoic acid. I'm not sure if this would still demonstrate freezing point depression since it is a colligative property based on amount of solute....

A compound containing only boron, nitrogen, and hydrogen was found to be 40.3% B, 52.2% N, and 7.5% H by mass. When 3.301 g of this compound is dissolved in 50.00 g of benzene, the solution produced freezes at 1.30 °C. The freezing point of pure benzene is 5.48 °C; Kb for benzene is 5.12 °C m¯1. What is the molecular weight of this compound?

Calculate the freezing point of a solution of 64.3 g of an ionic compound (FW = 136, i = 1.25) dissolved in 557 g of an organic solvent (FW = 96.22). The Kf of the solvent is 4.55 °C/m. The normal freezing point of the solvent is 11.1 °C. Report your answer to 1 decimal place. A Moving to another question will save this response. Ouortion 5 of 24

0.0851 kg of biphenyl (C H,) is dissolve in benzene (CH) to create a solution with a total volume of 350.0 mL. (Assume the change in volume is negligible) If the boiling point of pure benzene is 80.1 °C, then what would be the boiling point of this solution in °C? (Kb for benzene is 2.53 °C/m and the density of benzene is 0.877 g/mL)

The molecular weight of an organic compound was determined by measuring the freezing point depression of a benzene solution. A 0.500 g sample was dissolved in 50.0 g of benzene, and the resulting depression was 0.42°C (kf (benzene) = 5.065°C/m). What is the approximate molecular weight of the compound?

3. You are asked to calculate the Kf for t-butyl alcohol (MW = 74.122 g/mol). You gather the following information in your experiment. You weigh out 2.124 g t-butyl alcohol and you determine the freezing point of the solvent t-butyl alcohol is 21.6 °C (freezing point pure t-butyl alcohol). You then add water to the test tube with the t-butyl alcohol and measure the freezing point to be 15.3 °C. What key piece of information are you missing to calculate the Kf?