The molality of a solution $($denoted here by the symbol $m)$ is a measure of con$-$

centration, defined as moles of solute divided by mass of solvent, or in other

words

$m=\frac{n_{\text{m}\text{o}\text{l}\text{u}\text{t}\text{e}}}{M{n}_{\text{s}\text{o}\text{l}\text{v}\text{e}\text{n}\text{t}}}$

where $M$ is the solvent's molar mass. For aqueous solutions of NaCl at $25\mathrm{.}0C$

and $1$atm, and for $0m6mo\frac{l}{k}g,$ the available volumetric data can be fit

to the empirical function

$V=c_0+c_{1}m+c_2{m}^2+c_3{m}^3$

with parameters

$c_0=1001\mathrm{.}70c{m}^3$

$c_1=17\mathrm{.}298kgc\frac{m^3}{m}ol$

$c_2=0\mathrm{.}9777k{g}^{2}c\frac{m^3}{m}o{l}^2$

$c_3=-0\mathrm{.}0569k{g}^{3}c\frac{m^3}{m}o{l}^3.$

$($a$)$ Use this expression for $V$ to derive an expression for the partial molar volume

of NaCl.

$($b$)$ Determine the partial molar volume of water $($again in aqueoss NaCl $)$ as a

function of $m.$

$($c$)$ Given that the density of pure water $($at $25$ and $1$atm $)$ is $0\mathrm{.}99707\frac{g}{c}{m}^3,$

calculate the partial molar volume of water in a $1\mathrm{.}250$ molal solution of NaCl.

What is the fractional change in hat$(V{)}_{\text{w}\text{a}\text{t}\text{e}\text{r}}$ in this solution, relative to hat$(V{)}_{\text{s}\text{a}\text{t}\text{e}\text{r}}?$

$($d$)$ Given $1\mathrm{.}00kg$ of a $1\mathrm{.}250$ molal solution of aqueous NaCl, suppose that an

additional $10\mathrm{.}0g$ of NaCl is dissolved into this solution. Calculate the change

in the volume of the solution.

$($e$)$ What is the molar concentration $($moles of NaCl per liter of solution$)$ of a

solution that is $1\mathrm{.}250$ molal sodium chloride?