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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

A flask of ...

Question

A flask of $4.48$ L capacity contains a mixture of $N_{2}$ and $H_{2}$ at $0^{o}$ C and $1$ atm pressure. If the mixture is made to react to form $N H_{3}$ gas at the same temperature, the pressure in the flask reduces to $0.75$ atm. The partial pressure of $N H_{3}$ gas in the final mixture is?

0.33

atm

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0.50

atm

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0.66

atm

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0.25

atm

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Solution

The correct option is D $0.25$ atm
$N_{2}$ $+$ $3 H_{2} \to 2 N H_{3}$
Initially $a$ $a t m$ $b$ $a t m$ $0$
Finally $a - x$ $b - 3 x$ $2 x$

Thus, $a + b = 1$ and $a + b - 2 x = 0.75$
So, $2 x = 0.25 a t m$
and $P_{N H 3} = 2 x = 0.25 a t m$

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A flask of 4.48L capacity contains a mixture of N2 and H2 at 0oC and 1 atm pressure. If the mixture is made to react to form NH3 gas at the same temperature, the pressure in the flask reduces to 0.75atm. The partial pressure of NH3 gas in the final mixture is?

The correct option is D 0.25atm N2 + 3H2→2NH3Initially a atm b atm 0Finally a−x b−3x 2xThus, a+b=1 and a+b−2x=0.75So, 2x=0.25atmand PNH3=2x=0.25atm

A flask of $4.48$ L capacity contains a mixture of $N_{2}$ and $H_{2}$ at $0^{o}$ C and $1$ atm pressure. If the mixture is made to react to form $N H_{3}$ gas at the same temperature, the pressure in the flask reduces to $0.75$ atm. The partial pressure of $N H_{3}$ gas in the final mixture is?

The correct option is D $0.25$ atm
$N_{2}$ $+$ $3 H_{2} \to 2 N H_{3}$
Initially $a$ $a t m$ $b$ $a t m$ $0$
Finally $a - x$ $b - 3 x$ $2 x$

Thus, $a + b = 1$ and $a + b - 2 x = 0.75$
So, $2 x = 0.25 a t m$
and $P_{N H 3} = 2 x = 0.25 a t m$