Question

Equimolal solutions of NaCl and MgCl are prepared in water. Freezing point of NaCl is found to be - $2^{\circ }C$ what is the freezing point of MgCl${}_2$ solution?

Answer 1

Depression in freezing point= $i\times K_F\times m⟶\left(1\right)$

For $NaCl$ $i=2$ {$NaCl⟶N{a}^{+}+C{l}^{-}$}

$K_F$ for $H_{2}O=1.86deg\times C/m$

And ATQ, depression in freezing point of water= $2^{o}C$

or, $2=2\times 1.86\times m$

or, $m=0.538$ moles/Kg

Now, equimolar implies that

molality of $NaCl$= molality of $MgCl$

And for $MgCl$, $i=3$ {$MgC{l}_2⟶M{g}^{+2}+2Cl$}

Substituting the value in equation (1), we get,

Depression in freezing point= $3\times 1.86\times 0.538=3.002$

So, freezing point of $MgC{l}_2$ solution will be= $-{3.002}^{o}C$