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Q.
The partial pressure of hydrogen in a flask containing $2\, g \,H _{2}$ and $32\, g \,SO _{2}$ is
States of Matter
Solution:
Let total pressure be $P_{T}$
Then, $P _{ H _{2}}= P _{ T } \chi_{ H _{2}}$
Where, $\chi_{ H _{2}}=\frac{ n _{ H _{2}}}{ n _{ H _{2}}+ n _{ O _{2}}}$
$ n _{ H _{2}}=\frac{2\, g }{2\, g \,mol ^{-1}}=1 mol$
$n _{ SO _{2}}=\frac{32\, g }{64\, g\, mol ^{-1}}=\frac{1}{2} mol $
$\therefore n _{ H _{2}}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3} $
$P _{ H _{2}}=\frac{2}{3} P _{ T }$
i.e., $\frac{2}{3} rd$ of total pressure.