Q13.63 P - Chemistry: Molecular Nature Of Matter And Change

Question

How would you prepare the following aqueous solutions? 

(a) 57.5 mL of $\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}$ $\mathbf{Cr}{\left({\mathbf{NO}}_{\mathbf{3}}\right)}_{\mathbf{3}}$ from solid  $\mathbf{Cr}{\left({\mathbf{NO}}_{\mathbf{3}}\right)}_{\mathbf{3}}$.

(b) $\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}{\mathbf{m}}^{\mathbf{3}}$  of 1.45 M  ${\mathbf{NH}}_{\mathbf{4}}{\mathbf{NO}}_{\mathbf{3}}$from 2.50 M ${\mathbf{NH}}_{\mathbf{4}}{\mathbf{NO}}_{\mathbf{3}}$.

Step-by-Step Solution

Answer

1. 0.021g of Chromium nitrate should be dissolved in 57.5 mL of water to get $1.53\times {10}^{-3} M$  solution.
2. 2.50M of 3.4L $N{H}_{4}N{O}_3$ should be diluted to 5.8 L to get 1.45 M solution.

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.

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

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

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

Step 2: Expression to calculate the Concentration

1. Molarity of the  $\mathbf{Cr}{\left({\mathbf{NO}}_{\mathbf{3}}\right)}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}$

Volume of Solvent  $\mathbf{=}\mathbf{57}\mathbf{.}\mathbf{3}\mathbf{mL}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0573}\mathbf{L}$

$$\begin{gathered} \mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number}\mathbf{ } \mathbf{of}\mathbf{ } \mathbf{Moles} \mathbf{of} \mathbf{solute}\left(\mathbf{n}\right)}{\mathbf{Volume} \mathbf{of}\mathbf{ } \mathbf{the}\mathbf{ } \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}\mathbf{=}\frac{\mathbf{n}}{\mathbf{0}\mathbf{.}\mathbf{0573}\mathbf{L}} \\ \mathbf{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0573} \\ \mathbf{n}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{mol} \end{gathered}$$

The mass of $8.8\times {10}^{-5} mol Cr{\left(N{O}_3\right)}_3$  is:

 $$\begin{gathered} mass=n\times molar mass \\ =8.8\times {10}^{-5} mol\times 238.0 g/mol \\ =0.021 g \end{gathered}$$

b) Volume of Solvent  $\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}{\mathbf{m}}^{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{L}$

Molarity of  ${\mathbf{NH}}_{\mathbf{4}}{\mathbf{NO}}_{\mathbf{3}}$ at  $\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}{\mathbf{m}}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{45}\mathbf{M}$

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

Molarity of solute ${\mathbf{NH}}_{\mathbf{4}}{\mathbf{NO}}_{\mathbf{3}}$ at  ${\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{M}$

$$\begin{gathered} \mathbf{Molarity} \mathbf{Equation}\mathbf{,} \\ \mathbf{M}\mathbf{1}\mathbf{\times }\mathbf{V}\mathbf{1}\mathbf{=}\mathbf{M}\mathbf{2}\mathbf{\times }\mathbf{V}\mathbf{2} \\ 1.\mathbf{45}\mathbf{M}\mathbf{\times }\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{L}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{M}\mathbf{\times }\mathbf{V}\mathbf{2} \\ \mathbf{V}\mathbf{2}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{45}\mathbf{M}\mathbf{\times }\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{L}}{\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{M}} \\ \mathbf{V}\mathbf{2}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{L} \end{gathered}$$

2.50M of 3.4L  $N{H}_{4}N{O}_3$ should be diluted to 5.8 L to get 1.45 M solution.