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Verizon [17]
3 years ago
14

This question has multiple parts. Work all the parts to get the most points. An aqueous antifreeze solution is 39.0% ethylene gl

ycol () by mass. The density of the solution is 1.05 . a Calculate the molality of the ethylene glycol. Molality = mol/kg b Calculate the molarity of the ethylene glycol. Molarity = mol/L c Calculate the mole fraction of the ethylene glycol. Mole fraction =
Chemistry
1 answer:
nlexa [21]3 years ago
3 0

Answer:

Molality = <u>10.300 m</u>

<u>Molarity = 6.5970 M</u>

<u>mole fraction = </u>0.156549

Explanation:

39.0 % = ethylene glycol

61.0 % = water

imagine the total mass = 100g

39.0% ethylene glycol = 39g

61.0 % water = 61g

1) Molality = number of moles / mass of solvent (kg)

Molar mass of ethylene glycol = 62.07g/mole

mole = 39g / 62.07g/mole = 0.6283 moles

Molality = moles / mass of solvent = 0.6283 moles / 0.061kg = <u>10.300 m</u>

<u />

<u>2) Molarity</u> = number of moles / volume of solution

Since we know the density  of the solution = 1.05g /ml

⇒ volume = 100g / 1.05g /mL   = 95.24 mL = 0.09524 L

Molarity = 0.6283 moles / 0.09524 L = <u>6.5970 M</u>

3) Mole fraction

moles water = 61g / 18.02g/mole  = 3.38513 moles

Total number of moles = moles of ethylene glycol + moles of water = 0.6283 + 3.38513 = 4.0134276 moles

Mole fraction = 0.6283/ 4.0134276 = 0.156549

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B) 137 g NH₃

C) 2.30 g H₂

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Part B:

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Part C:

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Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

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3 years ago
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