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2 years ago

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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 =

1 answer:

nlexa [21]2 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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<u>Explanation:</u>

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Given mass of oxygen gas = 14.4 g

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2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = $\frac{2}{2}\times 0.45=0.45$ moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

$0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%$

Hence, the percent yield of the water is 31.98 %

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