- AFreezing point is lowered
- ✓Freezing point is raised
- CBoiling point is raised
- DBoth $(A)$ and $(C)$
Depression in freezing point and elevation in boiling point are directly proportional to the van 't hoff factor. Decreases in number of particles causes the increase in freezing point and decrease in the boiling point.
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(Atomic mass : $C =12 ; H =1 ; O =16$ )
$\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}\\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{C{H_3} - C - C{H_2}CO{C_2}{H_5}}
\end{array}$ will give iodoform test
Reason : It does not contains $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,}\\
{\,\,\,\,{\kern 1pt} \,\,\,\,\,\,\,\,\,\,\,||\,\,\,}\\
{C{H_3} - C - }
\end{array}$ group
$[ R =$ gas constant, $F =$ Faraday constant, $T =$ Temperature $]$
$(A)$ For the reaction, $M (s)+2 H ^{+}(a q) \rightarrow H _2(g)+ M ^{2+}(a q)$, if $\frac{ dE _{c o l l}}{ dT }=\frac{ R }{ F }$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes are temperature independent).
$(B)$ The cell reaction, $Pt (s) \mid H _2(g, 1$ bar $)\left| H ^{+}(a q, 0.01 M ) \| H ^{+}(a q, 0.1 M )\right| H _2(g, 1 bar ) \mid Pt (s)$, is an entropy driven process.
$(C)$ For racemization of an optically active compound, $\Delta S >0$.
$(D)$ $\Delta S >0$, for $\left[ Ni \left( H _2 O \right)_6\right]^{2+}+3$ en $\rightarrow\left[ Ni ( en )_3\right]^{2+}+6 H _2 O$ (where en $=$ ethylenediamine).