Question

a metal, m, of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as mcl$_x$. a boiling point elevation experiment is performed to determine the subscript x, and therefore, the formula of the salt. a 43.2 g sample of the salt is dissolved in 100.0 g of water and the boiling point of the solution is found to be 378.39 k. find the formula of the salt. assume complete dissociation of the salt in solution.
formula:

Explanation:

Step1: Find boiling point elevation

Boiling point of pure water is 373.15 K.
$\Delta T_b = 378.39\ \text{K} - 373.15\ \text{K} = 5.24\ \text{K}$

Step2: Relate to molality (i=1+x)

Boiling point elevation formula: $\Delta T_b = i \cdot K_b \cdot m$. For water, $K_b = 0.512\ \text{K·kg/mol}$. $i = 1+x$ (dissociates into $1\ \text{M}^{x+} + x\ \text{Cl}^-$).
$m = \frac{\Delta T_b}{i \cdot K_b} = \frac{5.24}{(1+x) \cdot 0.512}\ \text{mol/kg}$

Step3: Calculate molar mass of $\text{MCl}_x$

Molar mass $M = 56 + 35.45x\ \text{g/mol}$. Molality $m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{43.2}{(56+35.45x) \cdot 0.100}\ \text{mol/kg}$

Step4: Equate and solve for x

Set the two expressions for m equal:
$$\frac{43.2}{0.100(56+35.45x)} = \frac{5.24}{0.512(1+x)}$$
Cross-multiply:
$43.2 \cdot 0.512(1+x) = 5.24 \cdot 0.100(56+35.45x)$
$22.1184(1+x) = 0.524(56+35.45x)$
$22.1184 + 22.1184x = 29.344 + 18.5758x$
$22.1184x - 18.5758x = 29.344 - 22.1184$
$3.5426x = 7.2256$
$x \approx 2$

Answer:

$\text{MCl}_2$ (or specifically, $\text{FeCl}_2$, since M has atomic mass 56 amu, iron)