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When 2.66 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to effuse as the same amount of CO2 under the same conditions of temperature and pressure. What is the molar mass of the phosphorus chloride? 1. 138 g/mol 2. 102 g/mol 3. 156 g/mol 4. 87.7 g/mol

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Edufirst

Answer:

  • 138g/mol

Explanation:

As per Graham's law of effusion and diffusion, the rates of effusion of two gases are inversely related to their molar masses:

         [tex]\dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}}[/tex]

Since rates and times are inversely related, you can write

          [tex]\dfrac{\text{time of effusion B}}{\text{time of effusion A}}=\sqrt{\dfrac{M_B}{M_A}}[/tex]

For our case

   [tex]\dfrac{\text{time of effusion Phosphorus chloride}}{\text{time of effusion }CO_2}=\sqrt{\dfrac{\text{MPhosphorus Chloride}}{M CO_2}[/tex]

The molar mass of CO₂ is 44.01g/mol

Subsituting:

        [tex]1.77=\sqrt{\dfrac{\text{MPhosphorus Chloride}}{44.01g/mol}[/tex]

Solving:

         

        [tex]M_{\text{Phosphorus chloride}}=(1.77)^2\times 44.01g/mol=138g/mol[/tex]

The molar mass of the phosphorus chloride is 138 g/mol.

  • The calculation is as follows;

[tex]\frac{rate\ of\ effusion A }{rate\ of\ effusion B} = \sqrt{\frac{M_B}{M_A} }[/tex]

Since rates and times are inversely related,

So

[tex]\frac{time\ of\ effusion B }{time\ of\ effusion A} = \sqrt{\frac{M_B}{M_A} }[/tex]

Now

[tex]= 1.77 \times 44.01 g/mol[/tex]

= 138 g/mol

Here

The molar mass of CO₂ is 44.01g/mol

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