1) $\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha$
Hence
$\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha<<1)$
$2)$ $\mathrm{m}_2=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=7.68^{\circ} \mathrm{C}$
3) $\mathrm{m}_3=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}^{\prime}=3.8^{\circ} \mathrm{C}$
4) $\mathrm{m}_4=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}_{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}^{\prime}=1.86^{\circ} \mathrm{C}$
As per NCERT, $\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{k} \cdot \mathrm{kg} \mathrm{mol}^{-1}$
$\mathrm{k}_{\mathrm{f}}{ }^{\prime}($ Benzene $)=5.12 \mathrm{k} \cdot \mathrm{kg} \mathrm{mol}^{-1}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(i)$ $\begin{gathered}
HCN\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons {H_3}{O^ + }\left( {aq} \right) + C{N^ - }\left( {aq} \right) \hfill \\
{K_a} = 6.2 \times {10^{ - 10}} \hfill \\
\end{gathered} $
$(ii)$ $\begin{gathered}
C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
{K_b} = 1.6 \times {10^{ - 5}} \hfill \\
\end{gathered} $
These equilibria show the following order of the relative base strength
$Pt\,|\,B{r_{2(\ell )0.1M}}\,|\,Br_{({\text{aq}}),0.1M}^ - \,||\,H_{_{({\text{aq}})0.1M}}^ + \,|\,{H_{2(g)1atm}}\,|\,Pt$ given ; $E_{B{r^ - }/B{r_2}}^0\, = \, - 1.06\,V$
$C{H_3} - \mathop {\mathop {CH - }\limits_{|\,\,\,\,\,\,\,} }\limits_{C{H_3}\,} C{H_2} - C{H_2} - Cl$