How do you prepare $\mathbf{\text{250g}}$ of$\mathbf{\text{0.150m}}$ aqueous ${\mathbf{\text{NaHCO}}}_{\mathbf{\text{3}}}$ ?

Molality is defined as the moles of solute dissolved in mass of solvent.

Consider the given data below.

The molality is $0.150\text{m}$.

Assume the mass of water is $1\text{kg}$.

Define the number of moles of ${\text{NaHCO}}_{\text{3}}$as below.

$\text{molality}=\frac{{\text{moles\hspace{0.17em}of\hspace{0.17em}NaHCO}}_{\text{3}}}{\text{mass\hspace{0.17em}of\hspace{0.17em}water}}$

$\begin{array}{rcl}{\text{moles\hspace{0.17em}of\hspace{0.17em}NaHCO}}_{\text{3}}& =& \text{molality}\times \text{mass\hspace{0.17em}of\hspace{0.17em}water}\\ & =& 0.150\text{mol}/\text{kg}\times 1\text{kg}\\ & =& 0.150\text{moles}\end{array}$

Now, calculate the mass of the solute as below.

$\begin{array}{rcl}{\text{Massof\hspace{0.17em}NaHCO}}_{\text{3}}& =& {\text{moles\hspace{0.17em}of\hspace{0.17em}NaHCO}}_{\text{3}}\times {\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}NaHCO}}_{\text{3}}\\ & =& 0.150\text{mol}\times 84\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\\ & =& 12.6\text{g}\end{array}$

$\begin{array}{rcl}\text{mass\hspace{0.17em}of\hspace{0.17em}water}& =& 1\text{kg}\\ & =& 1000\text{g}\end{array}$

Thus, define mass of solution as below.

$\begin{array}{rcl}\text{mass\hspace{0.17em}of\hspace{0.17em}solution}& =& 12.6{\text{\hspace{0.17em}g\hspace{0.17em}of\hspace{0.17em}NaHCO}}_{\text{3}}+1000\text{\hspace{0.17em}g\hspace{0.17em}of\hspace{0.17em}water}\\ & =& 1012.6\text{g}\end{array}$

Mass of ${\text{NaHCO}}_{\text{3}}$ to prepare $250\text{g}$of $\text{0.150m}$aqueous solution can be calculated as follow.

role="math" localid="1663664403030" $$\begin{gathered} \text{1012.6g\hspace{0.17em}of\hspace{0.17em}total\hspace{0.17em}solution\hspace{0.17em}contains}=1{\text{2.6\hspace{0.17em}gNaHCO}}_{\text{3}} \\ \text{1g\hspace{0.17em}of\hspace{0.17em}total\hspace{0.17em}solution\hspace{0.17em}contains\hspace{0.17em}}=\frac{\text{12.6g}}{\text{1012.6g}} \end{gathered}$$

$\begin{array}{rcl}\text{250g\hspace{0.17em}of\hspace{0.17em}solution\hspace{0.17em}contains\hspace{0.17em}}& =& \frac{\text{12.6\hspace{0.17em}g}}{\text{1012.6\hspace{0.17em}g}}\times \text{250\hspace{0.17em}g}\\ & =& \text{3.11\hspace{0.17em}g}\end{array}$

Now, amount of water required to prepare $\text{250\hspace{0.17em}g}$solution of $0.150\text{m}$aqueous is,

role="math" localid="1663664718207" $\begin{array}{rcl}{\text{NaHCO}}_{\text{3}}& =& 250\text{g}-3.11\text{g}\\ & =& 246.89\text{g}\end{array}$

Hence, to prepare $\text{250\hspace{0.17em}g}$of $0.150\text{m}$aqueous ${\text{NaHCO}}_{\text{3}}$$3.11\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ and $246.89\text{g}$of water is required.