Question e0588

Dec 8, 2014

The answer is $0.62 L$.

Dilution calculations allow us to determine how a solute's concentration is reduced in solution usually by adding more solvent to the mix. This can be expressed mathematically by

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where

${C}_{1}$ = initial concentration (or molarity );
${V}_{1}$ = initial volume;
${C}_{2}$ - final concentration (or molarity);
${V}_{2}$ - final volume;

We can determine the final concentration of $N a C l$ solution by

${C}_{2} = {V}_{1} / {V}_{2} \cdot {C}_{1} = \frac{150 \cdot {10}^{- 3} L}{2.5 L} \cdot 6.5 M = 0.39 M$

We know that molarity is defined as

C = n_(solute)/(V_(solution)#, which gives us ${n}_{s o l u t e} = C \cdot {V}_{s o l u t i o n} = 0.39 \cdot 2.5 = 0.975 m o l e s$ in $2.5 L$.

However, we were given ${m}_{N a C l} = 13.8 g$, which translates, knowing that $N a C l$'s molar mass is equal to $58.5 \frac{g}{m o l}$, into

$n = \frac{13.8 g}{58.5 \frac{g}{m o l}} = 0.24 m o l e s$ This is roughly $\frac{1}{4}$th of the number of moles present in $2.5 L$, so we would expect the final volume to be roughly $\frac{1}{4}$th of $2.5 L \to 0.63 L$.

$V = \frac{n}{C} = \frac{0.24 m o l e s}{0.39 M} = 0.62 L$, which is close to our estimate.