What is the freezing point of water made by dissolving $\mathrm{10.11~g}$ of magnesium chloride in $\mathrm{86.03~g}$ of water?

The freezing-point depression constant of water is $\mathrm{1.86~^\circ C~m^{-1}}$.
To start I take the $\frac{10.11~\text{g}\ \ce{MgCl}}{86.03 \times 0.001 \times 1.86~\mathrm{^\circ C~m^{-1}})}$, which I keep getting as $\mathrm{63.18~^\circ C}$.

What am I doing wrong?

  • 1
    $\begingroup$ How did you manage to move the temperature from the bottom to the top? Oh wait, your equation is wrong. It should be $-1.86°/molar \times {10.11 g MgCl_2 / 95.21 g/mol \over 86.03 g \times 1kg/1000g}$. $\endgroup$ – LDC3 May 8 '15 at 3:25

The equation for freezing point depression is this:

$ΔTf=−K_f\cdot m \cdot i$.

Where $ΔT_f$ is the freezing point depression, $K_f$ is the freezing point depression constant $m$ is the molality of the solute, and $i$ is the van 't Hoff factor.

The van 't Hoff factor is the number of moles of ions the solute dissolves into, which in the case of $\ce{MgCl2 -> Mg^2+ + 2Cl-}$, is 3.

Your problem here is that you used the mass of $\ce{MgCl2}$ ($\ce{MgCl}$ is incorrect) instead of its molality, and did not include the van 't Hoff factor.

Molality is calculated by dividing moles of solute by kilograms of solvent. In this case,

$\frac{0.1062 \: \mathrm{mol \:of} \: \ce{MgCl2}}{0.08603 \: \mathrm{kg \: of} \: \ce{H2O}}$.

This shows that the molality of $\ce{MgCl2}$ is $1.234 \: \mathrm{mol~kg^{-1}}$. If you use this value instead of mass to calculate $ΔT_f$, the value comes out to be $-6.888 \:\mathrm{°C}$.


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