Q13.62 P - Chemistry: Molecular Nature Of Matter And Change

Question

How would you prepare the following aqueous solutions? 

(a) 1.40 L of 0.288 M KBr from solid KBr 

(b) 255 mL of 0.0856 M  ${\mathbf{LiNO}}_{\mathbf{3}}$ from 0.264 M ${\mathbf{LiNO}}_{\mathbf{3}}$ .

Step-by-Step Solution

Answer

 

1. 47.6 g of solid KBr should be dissolved in 1.40 L.
2. 0.264 M and 83 mL $LiN{O}_3$ solution should be diluted to 255 mL solution with water.

Step 1: Concentration of the Solution

Solution can be formed from the dissolution of the solute in the solvent. The solute decides the nature of the solution.

Molarity can be defined as the ratio of the mass of the solute to the volume of the solution present in litre measure.

$\mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number}\mathbf{ } \mathbf{of}\mathbf{ } \mathbf{Moles}}{\mathbf{Volume} \mathbf{of}\mathbf{ } \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}}$

Number of moles can be defined as the ratio of the mass of the atom/molecule and molar mass of the atom/molecule.

$\mathbf{Number} \mathbf{of}\mathbf{ } \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}}$

Step 2: Expression to calculate the Concentration

1. Molarity of the $\mathbf{KBr}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{288}\mathbf{M}$

          Volume of Solvent $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{L}$

$$\begin{gathered} \mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number}\mathbf{ } \mathbf{of} \mathbf{Moles}\left(\mathbf{n}\right)}{\mathbf{Volume}\mathbf{ } \mathbf{of} \mathbf{the}\mathbf{ } \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{0}\mathbf{.}\mathbf{288}\mathbf{M}\mathbf{=}\frac{\mathbf{n}}{\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{L}} \\ \mathbf{n}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{288} \mathbf{M}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{4} \mathbf{L} \\ \mathbf{n}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{mol} \end{gathered}$$

The mass of 0.40 mol of KBr is:

 $$\begin{gathered} Mass of KBr=0.40 mol\times 119.00 g/mol \\ =47.6 g \end{gathered}$$

Thus, 47.6 g of solid KBr should be dissolved in 1.40 L.

b) Volume of Solvent  $\mathbf{=}\mathbf{255}\mathbf{mL}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{255}\mathbf{L}$

Molarity of  ${\mathbf{LiNO}}_{\mathbf{3}}$ at  $\mathbf{0}\mathbf{.}\mathbf{255}\mathbf{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0856}\mathbf{M}$

Volume of Solvent  $\mathbf{=}{\mathbf{V}}_{\mathbf{2}}$

Molarity of solute  ${\mathbf{LiNO}}_{\mathbf{3}}$ at  ${\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{264}\mathbf{M}$

 $$\begin{gathered} \mathbf{Molarity} \mathbf{Equation}\mathbf{,} \\ {\mathbf{M}}_{\mathbf{1}}\mathbf{\times }{\mathbf{V}}_{\mathbf{1}}\mathbf{=}{\mathbf{M}}_{\mathbf{2}}\mathbf{\times }{\mathbf{V}}_{\mathbf{2}} \\ \mathbf{0}\mathbf{.}\mathbf{0856}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{255}\mathbf{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{264}\mathbf{M}\mathbf{\times }{\mathbf{V}}_{\mathbf{2}} \\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{0856}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{255}\mathbf{L}}{\mathbf{0}\mathbf{.}\mathbf{264}\mathbf{M}} \\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{083}\mathbf{L} \end{gathered}$$

0.264 M and 83 mL $LiN{O}_3$ solution should be diluted to 255 mL solution with water.