Question

$0.0852g$ of an organic halide $(A)$ when dissolved in $2.0g$ of camphor, the melting point of the mixture was found to be $167^{°}C$. Compound $(A)$ when heated with sodium gives a gas $(B)$. $280mL$ of gas $(B)$ at $STP$ weighs $0.375g$. What would be ${}^{\prime }{A}^{\prime }$ in the whole process? $K_f$ for camphor $=40$, $m$.$pt$. of camphor $=179^{°}C$.

• A. $C_2{H}_{5}Br$
• B. $C{H}_{3}I$
• C. $(C{H}_3{)}_{2}CHI$
• D. $C_3{H}_{7}Br$

Answer 1

Answer: (B)

$\Delta T=179-167=12$,
$w=0.0852g$,
$W=2g$,
$K_f=40$,
molecular weight of $\left(A\right)$
$=\frac{1000\times K_f\times w}{\Delta T\times W}$
$=\frac{1000\times 40\times 0.0852}{12\times 2}$
$=142$
$(A)$ undergoes Wurtz reaction to form $(B)$ i.e.
$(A)\stackrel{Na}{\to }(B)+NaX$
$(B)$ is an alkane say $C_n{H}_{2n+2}$
$\because 280mL$ of $(B)$ weighs $0.375g$ at $NTP$
$\therefore 22400mL$ of $(B)$ weighs $=\frac{0.375\times 22400}{280}$
$=30g$ at $NTP$
$\therefore$ M.wt. of $(B)=30$, $12n+2n+2=30$, $n=2$
Thus $(B)$ is ethane and therefore $(A)$ is $C{H}_{3}X$.
The m.wt. of $C{H}_{3}X=142$
At. wt. of $X=127\therefore X$ is iodine
Therefore alkyl halide is $C{H}_{3}I$. This reaction is
$\underset{\text{(A)}}{2C{H}_{3}I}\underset{\text{ether}}{\overset{Na}{\to }}\underset{\text{(B)}}{C_2{H}_6}$