When 12 . 2g of benzoic acid is dissolved in 100g of water, the...

${\Delta T}_{f} = i \times {k}_{f} \times m$

$0 - \left(- 0 . 93\right) = i \times 1 . 86 \times \dfrac{12 . 2}{122 \times 100} \times 1000$

$i = \dfrac{0 . 93}{1 . 86} = 0 . 5$

$i = 1 + \left(\dfrac{1}{n} - 1\right)\alpha$

$\dfrac{1}{2} = 1 + \left(\dfrac{1}{n} - 1\right) \times 1$

$n = 2$