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Standard XII

Chemistry

Dalton's Law of Partial Pressures

At 627 o C ...

Question

At $627^{o} C$ $1.0 a t m$ , ${S O}_{3}$ partially dissociates into ${S O}_{2}$ and $O_{2}$ . One liter of the equilibrium mixture weight $0.94 g$ under the above conditions. Determine the partial of the constituent gases in the mixture.

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Solution

Given $2 S O_{3} ⟷ 2 S O_{2} + O_{2}$

Weight of equilibrium mixture is $0.94$ g

Mol weight of equilibrium = sum of all molecules $= 176$ $g$

We know that $P V = n R T$

$P = \frac{W R T}{M V}$

$P = \frac{0.94}{176} \times \frac{0.0821 \times 900}{1} = 0.3946$ $a t m$

Hence partial pressure of gases is $0.3946$ $a t m$

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At 627oC 1.0atm, SO3 partially dissociates into SO2 and O2. One liter of the equilibrium mixture weight 0.94g under the above conditions. Determine the partial of the constituent gases in the mixture.

Given 2SO3⟷2SO2+O2 Weight of equilibrium mixture is 0.94g Mol weight of equilibrium = sum of all molecules =176 g We know that PV=nRT P=WRTMV P=0.94176×0.0821×9001=0.3946 atm Hence partial pressure of gases is 0.3946 atm

At $627^{o} C$ $1.0 a t m$ , ${S O}_{3}$ partially dissociates into ${S O}_{2}$ and $O_{2}$ . One liter of the equilibrium mixture weight $0.94 g$ under the above conditions. Determine the partial of the constituent gases in the mixture.