Q142CP - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

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.

A concept:

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 $1\text{ kg}$.

Moles of NaHCO 3 :

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.

$\begin{array}{rcl}{\text{Mass of \hspace{0.17em}NaHCO}}_{\text{3}}& =& {\text{moles\hspace{0.17em}of\hspace{0.17em}NaHCO}}_{\text{3}}\times {\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}NaHCO}}_{\text{3}}\\ & =& 0.150\text{ mol}\times 84\raisebox{1ex}{${\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 NaHCO 3 :

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

$$\begin{gathered} \text{1012.6 g \hspace{0.17em}of\hspace{0.17em}total\hspace{0.17em}solution\hspace{0.17em}contains}=1{\text{2.6\hspace{0.17em}g NaHCO}}_{\text{3}} \\ \text{1 g\hspace{0.17em}of\hspace{0.17em} total\hspace{0.17em}solution\hspace{0.17em}contains\hspace{0.17em}}=\frac{\text{12.6 g}}{\text{1012.6 g}} \end{gathered}$$

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

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

$\begin{array}{rcl}{\text{NaHCO}}_{\text{3}}& =& 250\text{ g}-3.11\text{ g}\\ & =& 246.89\text{ g}\end{array}$ 

 

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.