Question

Calculate the amount of \[KCl\] which must be added to $1\,Kg$ of water so that the freezing point is depressed by $2\,K$. \[(\,{k_f}\] for water $ = \,1.86\,K\,Kg\,mo{l^{ - 1}}\,)$

Answer 1

Hint:Freezing-point depression is the decrease of the freezing point of a solvent on the addition of a non-volatile solute and is given by the equation $\Delta {T_f} = i \times {k_f} \times m$where $i$is the Van’t Hoff Factor, ${k_f}$is the freezing point depression constant and $m$represents the molality of the solution.

Complete step-by-step answer:We know that the depression of freezing point is given by the equation $\Delta {T_f} = i \times {k_f} \times m........\left( 1 \right)$where $i$is the Van’t Hoff Factor, ${k_f}$is the freezing point depression constant and $m$represents the molality of the solution.
Now, the solution must have a solute and a solvent and in the given question $KCl$ is the solute and water is the solvent. $KCl$ undergoes dissociation in water as: $KCl\, \to \,{K^ + } + C{l^ - }$ and hence produces two ions in solution. So the Van’t Hoff Factor $\left( i \right)$ for the dissociation of $KCl$ is $2$.
${k_f}$ or the freezing point depression constant for water $ = \,1.86\,K\,Kg\,mo{l^{ - 1}}$.
$\Delta {T_f}\, = \,2\,K$.
Putting all the values in equation $\left( 1 \right)$ we get,
$2\,K\, = \,2 \times 1.86\,K\,Kg\,mo{l^{ - 1}} \times \,m$.
$ \Rightarrow \,m\, = \,\dfrac{{2\,K}}{{2 \times 1.86\,K\,Kg\,mo{l^{ - 1}}}}\, = \,0.538\,mol\,K{g^{ - 1}}$.
Now$m = \dfrac{{No.\,of\,moles\,of\,solute}}{{Mass\,of\,solvent\,\left( {in\,kg} \right)}}$. Since the mass of solvent i.e. water $ = \,1\,Kg$, therefore $m = \,No.\,of\,moles\,of\,solute$.
Also, $No.\,of\,moles\,of\,a\,given\,compound\, = \,\dfrac{{Given\,Mass\,of\,the\,compound}}{{Molar\,Mass\,of\,the\,compound}}$.
Therefore, $m\, = \,\dfrac{{Given\,Mass\,of\,the\,compound}}{{Molar\,Mass\,of\,the\,compound}}$.
Molar Mass of $KCl$ $ = \,74.5\,g\,mo{l^{ - 1}}$.
Let the given mass of $KCl$ i.e. the amount of $KCl$ added be $W\,g$.
Therefore the amount of $KCl$ added $ = \,m \times Molar\,Mass\,of\,KCl\, = \,\left( {0.538 \times \,74.5\,} \right)\,g\, = \,40.08\,g$

Note:You should take proper care of the units for these types of questions. Do not mix up the units and make sure to do the conversions wherever necessary. Try to proceed with the calculations in a stepwise manner as much as possible as this reduces the chances of doing any error.