Salt of F and alkali metal → produces 2 ions → i = 2

Use the equation:

$\overline{){\mathbf{\u2206}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{i}}{\mathbf{\times}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{\times}}{\mathbf{m}}}\phantom{\rule{0ex}{0ex}}\mathbf{m}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{mol}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}\phantom{\rule{0ex}{0ex}}\mathbf{m}=\mathbf{}\frac{\mathbf{\u2206}{\mathbf{T}}_{\mathbf{f}}}{\mathbf{i}\mathbf{\times}{\mathbf{K}}_{\mathbf{f}}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{mol}\mathbf{}\mathbf{solute}}{\mathbf{100}\mathbf{}\overline{)\mathbf{g}}\mathbf{}\mathbf{\left(}{\displaystyle \frac{1x{10}^{-3}\mathrm{kg}}{1\overline{)g}}}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{(}\mathbf{0}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{)}\mathbf{\xb0}\overline{)\mathbf{C}}}{\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{86}\mathbf{}{\displaystyle \frac{\overline{)\xb0C}}{m}}\mathbf{)}}\phantom{\rule{0ex}{0ex}}\mathbf{mol}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0376}\mathbf{}\mathbf{moles}\phantom{\rule{0ex}{0ex}}$

A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations.

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