$\begin{array}{1 1} 0.5 \\ 1 \\ 2 \\ 3 \end{array} $

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Answer :0.5

Amount of naphthoic acid $=20\;g/172\;g\;mol^{-1}$

$\qquad=(20/172)mol$

Molality of naphthoic acid $=(20/172)\;mol/0.05kg=(400/172)\;mol \;kg^{-1}$

Using the expression $(- \Delta T_t) =i K_t m$ we get,

$i= \large\frac {(- \Delta T_t)}{K_t m}=\frac{(2K)}{(1.72 K \;kg mol^{-1} ) ((400 /172) mol \;kg^{-1})}$$=0.5$

$\quad= 0.5$

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