CChapter 13 Exercise 13.80-Enhanced-with Feedback Part A Constants / Periodic Ta

Question

CChapter 13 Exercise 13.80-Enhanced-with Feedback Part A Constants / Periodic Table You may want to reference (Pages 593-607 Section 13.6 while completing this problenm Calculate the freezing point of the solution. Express the answer using four significant figures. Calculate the freezing point and boiling point of a solution containing 8.55 g of ethylene glycol (C2Hg02) in 96.7 mL of ethanol. Ethanol hasa density of 0.789 g/em eC Part B Calculale the boiling point of the solution. Calculate your answer using three significant figures. Request Answer MacBook Air 0 2 3 4 6

Explanation / Answer

Mass= volume * density

Mass= 0.789*96.7= 76.296 gram =0.0763kg

Concentration of ethylene glycol=8.55g/62.07g/mol=0.138 mol

Molality= 0.138mol/0.0763kg= 1.81 m

Part A Freezing point depression i = 1 since it doesn't ionize

t=iKfm

t= 1.99c/m*1.81=3.602 c

New Tf=-114.6-3.602=-117.8 C

Part B Boling point elevation

t=iKbm

t= 1.19c/m*1.81m=2.154 c

New Tb= 78 4+2.154=80.6 C

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