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timofeeve [1]
3 years ago
14

If 26.2 grams of a pure compound contain 8.77 × 1022 molecules, what is the molecular weight of this compound? Answer in units o

f g/mol.
Chemistry
1 answer:
Nataly [62]3 years ago
3 0

Answer:

Mw = 179.845 g/mol

Explanation:

  • Mw [=] g/mol

∴ w = 26.2 g

∴ 1 mol = 6.02 E23 molecules.......Avogadro's number

⇒N° moles = 8.77 E22 molecules * ( mol / 6.02 E23 molecules ) = 0.146 mol

⇒ Mw = 26.2 g / 0.146 mol = 179.845 g/mol

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4 years ago
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Answer:

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Explanation:

<u>What is effusion?</u>

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

<u>Graham's law of effusion of </u><u>gas</u>

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

\boxed{ \frac{Rate_1}{Rate_2} =  \sqrt{ \frac{M_2}{M_1} } }

<u>Calculations</u>

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Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

\frac{Rate_1}{Rate_2} = 1.7

1. 7 =  \sqrt{ \frac{M_2}{28} }

Square both sides:

2.89  = \frac{M_2}{28}

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

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The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

Thus based on the above reasoning, its probable identity is Krypton.

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