Question

What is the molality of\[NaCl\] if the freezing point of a \[NaCl\] solution is \[ - {5.58^o}C\]

Answer 1

Hint: Freezing point refers to the temperature at which liquid and solid form of the same material/substance are said to be in equilibrium and possess the similar vapour pressure. Freezing point depression is the lowering of freezing point of the solvent with the addition of solute to a solvent.

Complete step by step answer:
The lowering of freezing point or freezing point depression is a colligative property which depends upon the number of the solute particles. As the number of particle increases, the freezing point decreases. When salt for e.g. \[NaCl\] is added to water, this indicates that dissolved foreign particles have been introduced into the water. Freezing point of the water keeps on lowering as more and more particles are appended till the point when salt stops getting dissolved. This effect is known as depression in freezing point. The dissociation of \[NaCl\] in aqueous solution is depicted below in the equation:
$NaCl \to N{a^ + } + C{l^ - }$
It is clear from the above chemical equation that\[NaCl\] yields two ions (i.e. one sodium and one chloride ion).
As we know, depression in freezing point is directly proportional to the molality of the added solute. The formula used is:
$\Delta {T_f} = i \times {K_f} \times m$
Where, $\Delta {T_f} = $ depression in freezing point
$i = $ Van’t Hoff factor
 ${K_f} = $ cryoscopic constant
$m = $molality
In the present case, values will be:
$\Delta {T_f} = {T_f} - {T_i} = - 5.58 - 0 = - {5.58^o}C$ (freezing point of a\[NaCl\] solution is \[ - {5.58^o}C\])
$i = 2$ (\[NaCl\] yields two ions)
${K_f} = {1.86^o}C.kg.mo{l^{ - 1}}$ (in aqueous solution, solvent is water)
Now, substituting all the values in the aforementioned formula, we get:
\[
\Delta {T_f} = - {K_f} \times i \times m \\
m = - \dfrac{{( - 5.58)}}{{1.86 \times 2}} = 1.5\dfrac{{mol}}{{kg}} \\
\]
Hence, \[1.5\dfrac{{mol}}{{kg}}\] is the molality of \[NaCl\] if the freezing point of \[NaCl\] solution is\[ - {5.58^o}C\]

Note: The more solute or salt added to water, more is the depression in freezing point. Another colligative property which works in the same manner is the elevation in boiling point i.e. if you add any salt to the water, the freezing point is lowered and its boiling point is increased. Elevation in boiling point of a solvent is inversely proportional to its freezing point.