please within five minutes

label Chemistry
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

An aqueous antifreeze solution is 31.0% ethylene glycol (C2H6O2) by mass. 

The density of the solution is 1.039 g/cm3.

Calculate the mole fraction of ethylene glycol.

Oct 22nd, 2017

C2H6O10

Atomic mass of C =12

Atomic mass of H =1

Atomic mass of O =16

Atomic mass of C2H6O10 = 2*12+6*1+16*2 =62

C2 H6 O10 =0.31*1.039= 0.322

Water = 1.039-0.322 =0.717

Moles of C2 H6 O10 =0.322/62 =0.00519

Moles of water = 0.717/18 =0.0398

Mole fraction of Glycol = 0.00519/(0.00519+0.0398) = 0.115

Mole fraction of water = 0.0398/(0.00519+0.0398) = 0.885


Dec 10th, 2014

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Oct 22nd, 2017
...
Oct 22nd, 2017
Oct 23rd, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer