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5.19     A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

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A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen.

So, let us assume the weight of dihydrogen in a total mixture weight of W be 20Wgrams and dioxygen be 80Wgrams.

Then, the number of moles of H_{2} will be, n_{H_{2}} = \frac{20W}{2} = 10W\ moles and the number of moles of O_{2}n_{O_{2}} = \frac{80W}{32} = 2.5W\ moles.

And given that the total pressure of the mixture is P_{Total} = 1\ bar.

Then we have a partial pressure of H_{2},

P_{H_{2}}=(Mol.\ fraction)_{H_{2}}\times P_{Total}

= \frac{n_{H_{2}}}{n_{H_{2}}+n_{O_{2}}}\times P_{Total}

= \frac{10W}{10W+2.5W}\times 1\ bar

=0.8\ bar

Therefore, the partial pressure of H_{2} is 0.8\ bar.

Posted by

Divya Prakash Singh

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