Name:osh Chemistry 106 Spring 2018 Southeast Technical Institute Exam 4: Take-ho

Question

Name:osh Chemistry 106 Spring 2018 Southeast Technical Institute Exam 4: Take-home portion (35 pts) Show all work for each of the following questions. Explain your answers to each question and show calculations where appropriate. Clearly indicate your final answers to each question. If additional paper is needed to show work, staple to this examm Due: April 11th at 1pm at the beginning of class 1. A 789 L solution was created by dissolving 82.9 g of ithium phosphate in 7.5 L of propanol. The density of propanol is 803.4 kg/m'. The Kt for propanol is 2.,00°C/m and the freezing point is 127 ??. a. What is the solute in this solution? (2pts) b. What is the solvent in this solution? (2pts) Proparel Express the concentration in mass/mass % (3pts) c. d. Express the concentration in mass/volume % 13pts) e Express the concentration in molality (3pts) f Using the information given above, what temperature will the solution freeze at? (4pts) 127°c 2 31 845 3

Explanation / Answer

VOLUME OF PROPANOL = 7.5 L

DENSITY OF PROPANOL = 803.4 Kg/m3

MASS OF THE PROPANOL = DENSITY x VOLUME

= 803.4Kg/m3 x 7.5L = 6025.5g= 6.0255kg (10-3 m3 = 1L)   

MASS/MASS % = (MASS OF SOLUTE x 100)/MASS SOLUTION

= 82.9x100/(82.9+6025.5)

= 1.3571

ALREADY GIVEN IN QUESTION 7.89L OF SOLUTION AND 82.9g OF LITHIUM PHOSPHATE

MASS/VOLUME % = (MASS OF SOLUTE(g) x100)/VOLUME OF SOLUTION(mL)

= 82.9x100/7890mL =1.0507

MOLALITY = MOLES OF SOLUTE/ kg OF SOLVENT

=(MASS/MOLECULAR MASS)/ kg OF SOLVENT

=(82.9/115.79)/ 6.0255kg = 0.1187 molal

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