91Ó°ÊÓ

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.

$\text{molality}=\frac{{\text{³¾´Ç±ô±ð²õ o´Ú N²¹±á°ä°¿}}_{\text{3}}}{\text{³¾²¹²õ²õ â¶Ä‹o´Ú w²¹³Ù±ð°ù}}$

$\begin{array}{rcl}{\text{³¾´Ç±ô±ð²õ o´Ú N²¹±á°ä°¿}}_{\text{3}}& =& \text{molality}×\text{³¾²¹²õ²õ â¶Ä‹o´Ú w²¹³Ù±ð°ù}\\ & =& 0.150\text{mol}/\text{kg}×1\text{kg}\\ & =& 0.150\text{moles}\end{array}$

Now, calculate the mass of the solute as below.

$\begin{array}{rcl}{\text{²Ñ²¹²õ²õ´Ç´Ú N²¹±á°ä°¿}}_{\text{3}}& =& {\text{³¾´Ç±ô±ð²õ o´Ú N²¹±á°ä°¿}}_{\text{3}}×{\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ o´Ú N²¹±á°ä°¿}}_{\text{3}}\\ & =& 0.150\text{mol}×84\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\\ & =& 12.6\text{g}\end{array}$

$\begin{array}{rcl}\text{³¾²¹²õ²õ o´Ú w²¹³Ù±ð°ù}& =& 1\text{kg}\\ & =& 1000\text{g}\end{array}$

Thus, define mass of solution as below.

$\begin{array}{rcl}\text{³¾²¹²õ²õ o´Ú s´Ç±ô³Ü³Ù¾±´Ç²Ô}& =& 12.6{\text{ g o´Ú N²¹±á°ä°¿}}_{\text{3}}+1000\text{ g o´Ú w²¹³Ù±ð°ù}\\ & =& 1012.6\text{g}\end{array}$

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" $$\begin{gathered} \text{1012.6²µâ€‹â¶Ä‰o´Ú â¶Ä‹t´Ç³Ù²¹±ô s´Ç±ô³Ü³Ù¾±´Ç²Ô c´Ç²Ô³Ù²¹¾±²Ô²õ}=1{\text{2.6 g±·²¹±á°ä°¿}}_{\text{3}} \\ \text{1²µâ€‰o´Ú​â¶Ä‰t´Ç³Ù²¹±ô s´Ç±ô³Ü³Ù¾±´Ç²Ô â¶Ä‹c´Ç²Ô³Ù²¹¾±²Ô²õ }=\frac{\text{12.6g}}{\text{1012.6g}} \end{gathered}$$

$\begin{array}{rcl}\text{250²µâ€‰o´Ú s´Ç±ô³Ü³Ù¾±´Ç²Ô c´Ç²Ô³Ù²¹¾±²Ô²õ }& =& \frac{\text{12.6 g}}{\text{1012.6 g}}×\text{250 g}\\ & =& \text{3.11 g}\end{array}$

Now, amount of water required to prepare $\text{250 g}$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 $\text{250 g}$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.