Question

Calculate the amount of sodium chloride (in g) which must be added to $1000\, mL$ of water so that its freezing point is depressed by $0.744\,K$. For water, $K_{f} = 1.86\, K/m$. Assume density of water to be $1\, g \,mL^{-1}$.

Answer 1

Mass of $NaCl, w_{2} = $?, Volume of water $= 1000 mL$,
$\Delta T_{f} = 0.744 K$, Density of water $= 1 g mL^{-1}$
So, mass of water $W_{1} = 1000 mL \times 1 g mL^{-1}$
$= 1000g= 1 kg$
$\Delta T_{f} = iK_{f} m =iK_{f} \times\frac{\left(W_{2}58.5\right)}{W_{1}} = \frac{iK_{f} W_{2}}{58.5 \times W_{1}}$
$W_{2} =\frac{\Delta T_{f} \times58.5\times1}{i \times K_{f}} = \frac{0.744 \times58.5}{2\times1.86} = 11.7 g$
($\because i$ for $NaCl =2$)
So, Mass of $NaCl$ required $ = 11,7g$.