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pshichka [43]
3 years ago
8

A solution is prepared from 26.7 g of an unknown compound and 116.2 g acetone, C3H6O, at 313 K. At this temperature the vapor pr

essure of pure acetone is 0.526 atm while that of the solution is 0.501 atm. What is the molecular weight of the unknown compound?
Chemistry
1 answer:
Cloud [144]3 years ago
4 0

Answer:

The molecular weight of the unknown compound is 267.7 g/mol

Explanation:

Lowering vapor pressure → Colligative property where the vapor pressure of solution is lower than vapor pressure of pure solvent

ΔP = P° . Xm

0.526 atm - 0.501 atm = 0.025atm

0.025 atm = 0.526 atm . Xm

Xm = 0.025 atm / 0.526 atm → 0.0475 (mole fraction)

Mole fraction = Moles of solute / Total moles (solute + solvent)

0.0475 = Moles of solute / Moles of solute + Moles of solvent

We determine the moles of solvent → 116.2 g . 1mol / 58 g = 2 moles

0.0475 = Moles of solute / Moles of solute + 2

0.0475 moles of solute + 0.095 = Moles of solute

0.095 = Moles of solute - 0.0475 moles of solute

0.095 / 0.9525 = Moles of solute → 0.0997 moles

Molas mass = g/mol → 26.7 g / 0.0997 mol = 267.7 g/mol

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3 years ago
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8 0
3 years ago
Read 2 more answers
How many grams of aluminum metal can be produced when 50.0 grams of aluminum chloride decompose? 2AlCl3 → 2Al + 3Cl2
Arisa [49]

Answer:

10.1 g of Al are formed

Explanation:

The reaction is:

2AlCl3 --> 2Al + 3Cl2

So 2 moles of aluminun chloride decompose into 2 moles of Al and 3 moles of chlorine.

Ratio is 2:2.

Let's convert the mass of salt into moles (mass / molar mass)

50 g / 133.34 g/mol = 0.374 moles.

As the ratio is 2: 2, if I have 0.374 moles of salt, I would produce the same amount of Al, just 0.374.

Let's convert the moles to mass

(Mol . molar mass)

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3 years ago
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