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worty [1.4K]
3 years ago
9

A vessel of volume 100ml contains 10% of oxygen and 90% of an unknown gas. The gases diffuses in 86 second through a small hole

of vessel.
If pure oxygen under similar
conditions and diffuses in 75 second, find the molecular weight of unknown gas?​
Chemistry
1 answer:
swat323 years ago
7 0

The molecular weight of unknown gas : 23.46 g/mol

<h3>Further explanation</h3>

Given

A vessel contains 10% of oxygen and 90% of an unknown gas.

diffuses rate of mixed gas = 86 s

diffuses rate of O₂ = 75 s

Required

the molecular weight of unknown gas (M)

Solution

The molecular weight of mixed gas :(M O₂=32 g/mol)

\tt 0.1\times 32+0.9\times M=3.2+0.9M

Graham's Law :

\tt \dfrac{r_{O_2}}{r_{mixed~gas}}=\sqrt{\dfrac{M_{mixed}}{M_{O_2}} }\\\\\dfrac{75}{86}=\sqrt{\dfrac{3.2+0.9M}{32} }\\\\0.76=\dfrac{3.2+0.9M}{32}\\\\24.32=3.2+0.9M\\\\21.12=0.9M\rightarrow M=23.46~g/mol

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Answer:

1520mmHg

Explanation:

Data obtained from the question include:

V1 (initial volume) = 600 mL

P1 (initial pressure) = 760 mmHg

V2 (final volume) = 300 mL

P2 (final pressure) =.?

Using the Boyle's law equation P1V1 = P2V2, the final pressure of the gas can easily be obtained as shown below:

P1V1 = P2V2

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AgNO₃ will act as the oxidising agent.

<h3><u>For the given chemical equation:</u></h3>

Cu + 2AgNO₃ →  2Ag + Cu(NO₃)₂

Half reactions for the given chemical reaction:

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Copper is a reducing agent because it is losing 2 electrons, which causes an oxidation process.

<u>Oxidising Agent</u>:

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The silver ion undergoes a reduction process and is regarded as an oxidizing agent since it is acquiring one electron per atom.

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Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 10 kPa

P_2 = final pressure of gas = 15 kPa

V_1 = initial volume of gas = 10m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 50^oC=273+50=323K

T_2 = final temperature of gas = 75^oC=273+75=348K

Now put all the given values in the above equation, we get:

\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}

V_2=7.2m^3

The new volume of carbon dioxide gas is 7.2m^3

Now we have to calculate the new density of carbon dioxide gas.

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Formula for new density will be:

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where,

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M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

\rho = new density

Now put all the given values in the above equation, we get:

\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}

\rho_2=0.2281g/L

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