The value of Van't Hoff factor (i) for K 3[ Fe ( CN ] is X, if it undergoes 75 % dissociation in aqueous medium, calculate 4 X ?

Question

The value of Van't Hoff factor (i) for K 3[ Fe ( CN ] is X, if it undergoes 75 % dissociation in aqueous medium, calculate 4 X ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-ss9yxdeupi3yqghb.jpg" /> i=(C - CÎ± + 3 CÎ± + CÎ±/C)=1+3α Given α =0.75 ∴ i=1+3× 0.75=3.254X=4× 3.25=13

Q. The value of Van't Hoff factor (i) for $K_{3} [F e (CN]$ is $X$ , if it undergoes $75%$ dissociation in aqueous medium, calculate $4 X$ ?

$i = \frac{C - C α + 3 C α + C α}{C} = 1 + 3 α$
Given $α = 0.75$
$∴ i = 1 + 3 \times 0.75 = 3.254 X = 4 \times 3.25 = 13$

Q. The value of Van't Hoff factor (i) for K3​[Fe(CN] is X, if it undergoes 75% dissociation in aqueous medium, calculate 4X ?

i=CC−Cα+3Cα+Cα​=1+3α Given α=0.75 ∴i=1+3×0.75=3.254X=4×3.25=13

Q. The value of Van't Hoff factor (i) for $K_{3} [F e (CN]$ is $X$ , if it undergoes $75%$ dissociation in aqueous medium, calculate $4 X$ ?

$i = \frac{C - C α + 3 C α + C α}{C} = 1 + 3 α$
Given $α = 0.75$
$∴ i = 1 + 3 \times 0.75 = 3.254 X = 4 \times 3.25 = 13$