\(\begin{array}{llll} t =0  a  0  0\end{array}\)

\(\begin{array}{llll} t = t  a - a \alpha  a \alpha  2 a \alpha\end{array}\)

\(n _{ T }= a - a \alpha+ a \alpha+2 a \alpha\)

\(= a (1+2 \alpha)\)

so \(i =1+2 \alpha\)

Now \(\Delta T _{ b }= i \times m \times K _{ b }\)

\(\Delta T _{ b }=(1+2 \alpha) \times m \times K _{ b }\)

\(\alpha=0.1 \quad m =10 \quad K _{ b }=0.5\)

\(\Delta T _{ b }=1.2 \times 10 \times 0.5\)

\(=6\)

So boiling point \(=106\)