Question

equal weight of ch4 and H2 are mixed in a container at 25celsius fraction of total pressure exerted by methane is
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Answer 1

Let x grams of each hydrogen and methane are mixed, Moles of H2 = x/2 Moles of CH4 = x/16 ⇒ Mole-fraction of H2 = (x/2)/(x/2 + x/16) = 8/9 ⇒ (Partial pressure of H_2)/(Total pressure) = mole-fraction of H2 = 8/9

Answer 2

Answer : The fraction of total pressure exerted by methane is, $\frac{1}{9}$

Explanation :

As per question, the mass of $CH_4$ and $H_2$ are equal. So, let the mass of $CH_4$ and $H_2$ is equal to 'M' grams.

First we have to calculate the moles of $CH_4$ and $H_2$.

$\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}=\frac{Mg}{16g/mole}=\frac{M}{16}moles$

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{Mg}{2g/mole}=\frac{M}{2}moles$

Now we have to calculate the mole fraction of $CH_4$ and $H_2$.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }H_2}$

$\text{Mole fraction of }CH_4=\frac{\frac{M}{16}}{\frac{M}{16}+\frac{M}{2}}=\frac{1}{9}$

$\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }CH_4+\text{Moles of }H_2}$

$\text{Mole fraction of }H_2=\frac{\frac{M}{2}}{\frac{M}{16}+\frac{M}{2}}=\frac{8}{9}$

Now we have to calculate the fraction of total pressure exerted by methane.

$P_{CH_4}=X_{CH_4}\times P_T$

where,

$P_{CH_4}$ = partial pressure of methane

$P_T$ = total pressure

$X_{CH_4}$ = mole fraction of methane

Now put all the given values in this formula, we get:

$P_{CH_4}=\frac{1}{9}\times P_T$

$\frac{P_{CH_4}}{P_T}=\frac{1}{9}$

Therefore, the fraction of total pressure exerted by methane is, $\frac{1}{9}$