MCQ
How will you separate a solution (miscible) of benzene $ + CHC{l_3}$
- ASublimation
- BFiltration
- ✓Distillation
- DCrystallisation
✓
Answer
Correct option: C.
Distillation
c
(c)Distillation particularly fractional distillation because the boiling point of benzene $({80\,^o}C)$ and chloroform $({61.5\,^o}C)$ are close.
(c)Distillation particularly fractional distillation because the boiling point of benzene $({80\,^o}C)$ and chloroform $({61.5\,^o}C)$ are close.
Fractional distillation involves repeated distillations and condensations, in a fractionating column. As a result of distillation and condensation at each point of the fractionating column, the vapours rising up become richer in more volatile component and the liquid falling back into the flask becomes richer in less volatile component. Thus, the low boiling liquid distils first while the higher boiling liquid distils afterwards.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The reaction giving incorrect major product
The spin-only magnetic moment value for the complex $\left[\mathrm{Co}\left(\mathrm{CN}_{6}\right)\right]^{4-}$ is $......\,BM$ [At. no. of $\mathrm{Co}=27$ ]
→The values of observed and calculated molecular mass of silver nitrate are $92.64$ and $170$ respectively. The degree of dissociation of silver nitrate is ........ $\%$.
→For a $d$ -electron, the orbital angular momentum is:
→The first order rate constant $K$ is related to temp. as $\log\, k = 15.0 - (10^6 /T)$ Which of the following pair of value is correct?
→Compound $A$ was treated with a large excess of $CH_3MgBr$. The resulting product was exposed to $POCl_3$ /pyridine to give compound $B$, as one of many products
→Which of the following compounds can be $A$ ?
The $3d-$ metal ions are paramagnetic in nature because
→In $C{H_3}C{H_2}OH\mathop {\xrightarrow{X}}\limits_{{{350}\,^o}C} C{H_2} = C{H_2} + {H_2}O;$ $'X' $ is
→An aromatic compound $'A'$ $C_7H_6Cl_2,$ gives $AgCl$ on bonding with alcoholic $AgNO_3$ solution, and yields $C_7H_7OCl$ on treatment with sodium hydroxide. $'A'$ on oxidation gives a mono chlorobenzoic acid which affords only one mononitro derivative. The compound $A$ is :
→Thermally most stable compound is