Question

The value of vant hoff factor for a soln of Bacl2 upon 50%dissociation is​

Answer 1

Answer:

$BaCl </p><p>2</p><p> </p><p> ⇌Ba </p><p>2+</p><p> +2Cl </p><p>−$

$Initially 1 0 0</p><p>After disso. 1−α α 2α$

$Total moles after dissociation</p><p></p><p>=1−α+α+2α</p><p></p><p>=1+2α$

$i= </p><p>normalmolesofsolute</p><p>observedmolesofsolute</p><p> </p><p> = </p><p>1</p><p>1+2α</p><p> </p><p> </p><p></p><p>∴α= </p><p>2</p><p>i−1</p><p> </p><p> = </p><p>2</p><p>1.98−1</p><p> </p><p> = </p><p>2</p><p>0.98</p><p> </p><p> =0.49$

$</p><p>For 1 mole α=0.49$

$For 0.01 mole α= </p><p>0.01</p><p>0.49</p><p> </p><p> =49%$