A mixture contains 16 grams of oxygen, 28 grams of nitrogen and 8 grams of methane. Total pressure of the mixture is 740 $m m$ . What is the partial pressure of nitrogen in $m m$ ?

185

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370

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555

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740

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Solution

The correct option is A 370
According to Dalton's Law of partial pressure
$p_{A} = χ_{A} {. P}_{t o t a l}$
where, $p_{A}$ = partial pressure of gas A in the mixture,
$χ_{A}$ = mole fraction of gas A $= \frac{N u m b e r o f m o l e s o f g a s^{'} A^{'} (n_{A})}{T o t a l n u m b e r o f m o l e s o f g a s e s p r e s e n t i n t h e m i x t u r e (n_{t o t a l})}$
$P_{t o t a l}$ =Total pressure of the mixture = 740 $m m$
Number of moles of gas A, $n_{A} = \frac{m a s s o f g a s^{'} A^{'} (m_{A})}{M o l a r m a s s o f g a s^{'} A^{'} (M_{A})}$
Using, $m_{O_{2}} = 16 g, M_{O_{2}} = 32 g / m o l$ , $m_{N_{2}} = 28 g, M_{N_{2}} = 28 g / m o l$ and $m_{{C H}_{4}} = 8 g, M_{{C H}_{4}} = 16 g / m o l$
Calculating number of moles of each gas, we get: $n_{O_{2}} = 0.5 m o l, n_{N_{2}} = 1 m o l a n d n_{{C H}_{4}} = 0.5 m o l$
Partial pressure of Nitrogen is therefore,
$p_{N_{2}} = χ_{N_{2}} P_{t o t a l} = \frac{1}{0.5 + 1 + 0.5} \times 740 m m = 370 m m$

Option B is the correct answer.

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