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Question: A mixture of hydrogen and oxygen at one bar pressure contains \[20\% \] by weight of hydrogen. Parti...

A mixture of hydrogen and oxygen at one bar pressure contains 20%20\% by weight of hydrogen. Partial pressure of hydrogen will be?
A. 0.2  bar{\rm{0}}{\rm{.2}}\;{\rm{bar}}
B. 0.4  bar{\rm{0}}{\rm{.4}}\;{\rm{bar}}
C. 0.6  bar{\rm{0}}{\rm{.6}}\;{\rm{bar}}
D. 0.8  bar{\rm{0}}{\rm{.8}}\;{\rm{bar}}

Explanation

Solution

When a gas is a mixture of different types, the total pressure is the sum of pressures exerted by each gas. Pressure added by linearly. The pressure added by each individual gas is called its partial pressure.

Complete step by step answer:
For an ideal solution and ideal gases, the partial pressure of a component will be directly proportional to its molar fraction.
Generally, the partial pressure can be calculated by multiplying the mole fraction of the gas with the total pressure of the container.
Mixture contains 20  %20\;\% by weight of hydrogen means 100100gram of the mixture contains 2020 gram of hydrogen, therefore 80 gram of the mixture will be oxygen. Now, let’s calculate the moles of hydrogen and oxygen as follows;
No.  of  moles  of  hydrogen =wt.mol.wt =202 =10  moles {\rm{No}}{\rm{.}}\;{\rm{of}}\;{\rm{moles}}\;{\rm{of}}\;{\rm{hydrogen}}\\\ = \dfrac{{{\rm{wt}}{\rm{.}}}}{{{\rm{mol}}{\rm{.wt}}}}\\\ = \dfrac{{20}}{2}\\\ = 10\;{\rm{moles}}
No.  of  moles  ofoxygen =wtmol.wt =8032 =2.5  moles {\rm{No}}{\rm{.}}\;{\rm{of}}\;{\rm{moles}}\;{\rm{of}}{\rm{oxygen}}\\\ = \dfrac{{{\rm{wt}}}}{{{\rm{mol}}{\rm{.wt}}}}\\\ = \dfrac{{80}}{{32}}\\\ = 2.5\;{\rm{moles}}
Since we know the relation between the mole fraction of hydrogen, mass of hydrogen and total number of moles. Therefore,

= \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{hydrogen}}}}{{{\rm{Total}}\;{\rm{no}}{\rm{.}}\;{\rm{of}}\;{\rm{moles}}}}$$ Substitute the values in the above formula. $ {\rm{Mole}}\;{\rm{fraction}} = \dfrac{{10}}{{12.5}}\\\ = 0.8 $ According to the formula,

{\rm{Partial pressure of hydrogen}} = {X_{{\rm{hydrogen}}}} \times {\rm{pure}};{\rm{pressure}}\\
0.8 \times 1\\
= 0.8;{\rm{bar}}

∗∗Therefore,thecorrectoptionisoption(D).∗∗∗∗Note:∗∗Dalton’slawofpartialpressuretellsusthatthetotalpressureofasystemofgasesisequaltothetotalsumofpartialpressureofeachofthedifferentgases.Dalton’slawholdsformixturesofgasesorpuregases. **Therefore, the correct option is option (D).** **Note:** Dalton’s law of partial pressure tells us that the total pressure of a system of gases is equal to the total sum of partial pressure of each of the different gases. Dalton’s law holds for mixtures of gases or pure gases.