The molecular mass of PCl5 is 208-32 but when heated to 230-C- it is reduced to 124- The extent of dissociation of PCl5 at this tempearature will be

Α = (M M_e)/M_e(y 1) = 208.32 124/124 × (2 1) = 0.68 y = 2 for PCl_5 ⇌ PCl_3 + Cl_2 ...

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Standard XII

Chemistry

Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

The molecular...

Question

The molecular mass of $P C l_{5}$ is 208.32 but when heated to $230 \circ C$ , it is reduced to 124. The extent of dissociation of $P C l_{5}$ at this tempearature will be

6.8%

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46%

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64%

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68%

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Solution

The correct option is D
68%

$α = \frac{(M - M_{e})}{M_{e} (y - 1)} = \frac{208.32 - 124}{124 \times (2 - 1)} = 0.68$
$y = 2 f o r P C l_{5} ⇌ P C l_{3} + C l_{2}$

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Similar questions

The molecular mass of $P C l_{5}$ is 208.32 but when heated to $230 \circ C$ , it is reduced to 124. The extent of dissociation of $P C l_{5}$ at this tempearature will be

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The molecular mass of PCl5 is 208.32 but when heated to 230∘C, it is reduced to 124. The extent of dissociation of PCl5 at this tempearature will be

The correct option is D 68% α=(M−Me)Me(y−1)=208.32−124124×(2−1)=0.68 y=2 for PCl5⇌PCl3+Cl2

The molecular mass of $P C l_{5}$ is 208.32 but when heated to $230 \circ C$ , it is reduced to 124. The extent of dissociation of $P C l_{5}$ at this tempearature will be

The correct option is D
68%

$α = \frac{(M - M_{e})}{M_{e} (y - 1)} = \frac{208.32 - 124}{124 \times (2 - 1)} = 0.68$
$y = 2 f o r P C l_{5} ⇌ P C l_{3} + C l_{2}$