$\frac{\Delta T _{ b }}{\Delta T _{ f }}=\frac{ K _{ b } \times 1}{ K _{ f } \times 2}$ $\Rightarrow \frac{3}{6}=\frac{1}{2}=\frac{ K _{ b }}{ K _{ f }} \times \frac{1}{2}$
$\frac{ K _{ b }}{ K _{ f }}=\frac{1}{1} \Rightarrow x =1$
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$A \to {\text{Product ;}}\,\, - \frac{{d[A]}}{{dt}} = {k_1}{[A]^o}$
$B \to {\text{Product ;}}\,\, - \frac{{d[B]}}{{dt}} = {k_2}{[B]}$
Units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as
$3HCHO+A\xrightarrow[{{40}\,^{o}}C]{N{{a}_{2}}C{{O}_{3}}}\underset{(82\%)}{\mathop{(B)}}$
Product $(B)$ of the above reaction is
$C{{H}_{3}}CN+2H\underset{\text{Ether}}{\mathop{\xrightarrow{HCl}}}\,X\xrightarrow{\text{Boiling }{{H}_{2}}O}Y;$ the term Y is