Question

Mole fraction of O2 = 571 Correct The mole fraction is the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. The mole fractions can be represented in terms of pressures The total pressure PTOTAL = 0.180 atm +0.240 atm = 0.420 atm The mole fraction of CH4 in the mixture PCH 0.180 atm 0.420 atm = 0.429 PTOTAL Since it is the mixture of two gases, the mole fraction of O, can be determined by subtraction Xo, =1.000 – 0.429 = 0.571 D If the mixture occupies a volume of 10.2 L at 61°C, calculate the total number of moles of gas in the mixture. Total number of moles = 0.156 mol

Answer 1

Hi,

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I guess you need the answer for PART (c):

We have, mole fraction is given by,

$X=\frac{no.of\;moles}{n_{total}}$

Given that,

Total no.of moles, $n_{total}=0.156\;mol$

Mole fraction of CH₄, $X_{CH_{4}}=0.429$

Hence, No.of moles of CH₄ is given by,

$n_{CH_{4}}=X_{CH_{4}}\times n_{total}=0.429\times0.156\;mol$

$\therefore n_{CH_{4}}=0.066924\;mol$

We know that, molar mass of CH₄ is given by, $M=12\;g/mol+4\times(1\;g/mol)=16\;g/mol$

Hence, mass of CH₄ is:

$m_{CH_{4}}=M\times n_{CH_{4}}=16\;g/mol\times0.066924\;mol$

$\mathbf{\therefore m_{CH_{4}}=1.07\;g}$

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Similarly, for oxygen:

Given that,

Total no.of moles, $n_{total}=0.156\;mol$

Mole fraction of O₂, $X_{O_{2}}=1-X_{CH_{4}}=1-0.429=0.571$

Hence, No.of moles of O₂ is given by,

$n_{O_{2}}=X_{O_{2}}\times n_{total}=0.571\times0.156\;mol$

$\therefore n_{O_{2}}=0.089076\;mol$

We know that, molar mass of O₂ is given by, $M=32\;g/mol$

Hence, mass of O₂ is:

$m_{O_{2}}=M\times n_{O_{2}}=32\;g/mol\times0.089076\;mol$

$\mathbf{\therefore m_{O_{2}}=2.85\;g}$

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