MCQ
The decreasing order of boiling points is
- ✓$n-$ Pentane > iso-Pentane > neo-Pentane
- Biso-Pentane > $ n-$ Pentane > neo-Pentane
- Cneo-Pentane > iso-Pentane >$ n-$ Pentane
- D$n-$ Pentane > neo-Pentane > iso-Pentane
So, higher the branches, lower is the boiling point.
So, the order of boiling point of isomeric pentanes is $n -$ pentane $\,>\,$ iso $-$ pentane $\,>\,$ neo $-$ pentane.
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Given : Atomic masses of $C , H$ and $N$ are $12,1$ and $14\, amu$ respectively.
The molar mass of the compound $A$ is $162\, g\, mol ^{-1}$.
$I.$ $1$ molecule of oxygen
$II.$ $1$ atom of nitrogen
$III.$ $1 \times {10^{ - 10}}$ $g$ molecular weight of oxygen
$IV.$ $1 \times {10^{ - 10}}$ $g$ atomic weight of copper
| List $-I$ | List $-II$ |
|
$A.$ Melting point $[\mathrm{K}]$ |
$I.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$ |
|
$B.$ Ionic Radius $\left[\mathrm{M}^{+3} / \mathrm{pm}\right]$ |
$II.$ $\mathrm{B}>\mathrm{Tl}>\mathrm{Al} \approx \mathrm{Ga}>\mathrm{In}$ |
| $C.$ $\Delta_{\mathrm{i}} \mathrm{H}_1 $ $ [\mathrm{~kJ} \mathrm{~mol}^{-1}]$ | $III.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}>\mathrm{B}$ |
| $D.$ Atomic Radius $[pm]$ | $IV.$ $\mathrm{B}>\mathrm{Al}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}$ |