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2 years ago

A 0.465 g sample of an unknown compound occupies 245 mL at 298 K and 1.22 atm. What is the molar mass of the unknown compound?

1 answer:

5 0

Using the Ideal Gas Law
PV= nRT
PV = (m/Mr) RT The molar mass of the unknown compound is : 38.0 g/mol

Mr = mRT / PV = 0.465 g x 0.08206 L * atm * K⁻¹ mol ⁻¹ x 298 K / 1.22 atm x 0.245 L = 38.0 g / mol

everything else but G / Mol are cancelled out.

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Explanation:

What is effusion?

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

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• 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

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

Calculations

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

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)

$\frac{Rate_1}{Rate_2} = 1.7$

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$2.89 = \frac{M_2}{28}$

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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.

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

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