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aalyn [17]
3 years ago
7

An aqueous antifreeze solution is 70.0% ethylene glycol (c2h6o2) by mass. the density of the solution is 1.75 g/cm3. calculate t

he molality, molarity, and mole fraction of the ethylene glycol.
Chemistry
1 answer:
AVprozaik [17]3 years ago
6 0

Molarity = moles of solute / kg solvent

 

Assume you have exactly 1 kg of solution. 70% of that, or 700 grams of the solution is C2H6O2, and 300 g of the solution is water.

 

Moles C2H6O2 = 700 g / 62.07 g/mol = 11.28 mol C2H6O2

 

molarity = 11.28 mol / 0.3 kg H2O = 37.59 molal

__________________________

 

Molarity = moles of solute/L of solution

there is 1 kg of solution has a volume of:

 

1000 g / 1.05 g/mL = 952 mL = 0.952 L

 

Molarity = 11.28 mol / 0.952 L = 11.85 M

___________________________

 

Mole fraction ethylene glycol = moles ethylene glycol / (moles ethylene glycol + moles H2O)

 

moles ethylene glycol = 11.28

moles H2O = 300 g / 18.0 g/mol = 16.67 mol H2O

 

mole fraction = 11.28 /( 11.28 + 16.67) = 0.404

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Answer:

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d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Explanation:

a) AgNO3 + KI → Ag+ + NO3- + K+ + I-

Ag+ + NO3- + K+ + I-  → AgI + KNO3

AgNO3 + KI → AgI + KNO3

b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-

Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O

Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O

c) 2Na3PO4  + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-

6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- →  Ni3(PO4)2 + 6NaNO3

2Na3PO4  + 3Ni(NO3)2  → Ni3(PO4)2 + 6NaNO3

d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-

2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

7 0
3 years ago
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Answer:

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