Q13.61 - Chemistry: Molecular Nature Of Matter And Change

Question

How would you prepare the following aqueous solutions? 

(a) 2.5 L of 0.65 M NaCl from solid NaCl 

(b) 15.5 L of 0.3 M urea $\mathbf{[}{\left({\mathbf{NH}}_{\mathbf{2}}\right)}_{\mathbf{2}}\mathbf{C}\mathbf{=}\mathbf{O}\mathbf{]}$ from 2.1 M urea.

Step-by-Step Solution

Answer

a) 9.53 g of solid NaCl should be added to 2.5 L of water.

b) 2.2L of urea with concentration 2.1 M should be diluted to 15.5 L.

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.

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

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

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

Step 2: Expression to calculate the Concentration

1. Molarity of the NaCl is 0.65 M.

          Volume of Solution 2.5 L.

$$\begin{gathered} \mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Volume} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{0}\mathbf{.}\mathbf{065}\mathbf{=}\frac{\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}}{\mathbf{2}\mathbf{.}\mathbf{5}} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{065}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{5} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{163}\mathbf{moles} \end{gathered}$$

The mass of NaCl is:

  $$\begin{gathered} Mas{s}_{NaCl}=0.163 mol\times 58.5 g/mol \\ =9.53 g \end{gathered}$$

Hence, 9.53 g of NaCl should be added to 2.5 L of water.

b) Volume of Solvent  $\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{L}$

Molarity of urea  $\mathbf{[}{\left({\mathbf{NH}}_{\mathbf{2}}\right)}_{\mathbf{2}}\mathbf{C}\mathbf{=}\mathbf{O}\mathbf{]}$ at  $\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{M}$

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

Molarity of solute urea  $\mathbf{[}{\left({\mathbf{NH}}_{\mathbf{2}}\right)}_{\mathbf{2}}\mathbf{C}\mathbf{=}\mathbf{O}\mathbf{]}$ at  ${\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{M}$

$$\begin{gathered} \mathbf{Molarity}\mathbf{ } \mathbf{Equation}\mathbf{,} \\ {\mathbf{M}}_{\mathbf{1}}\mathbf{\times }{\mathbf{V}}_{\mathbf{1}}\mathbf{=}{\mathbf{M}}_{\mathbf{2}}\mathbf{\times }{\mathbf{V}}_{\mathbf{2}} \end{gathered}$$

$$\begin{gathered} \mathbf{0}\mathbf{.}\mathbf{3}\mathbf{M}\mathbf{\times }\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{L}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{M}\mathbf{\times }{\mathbf{V}}_{\mathbf{2}} \\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{M}\mathbf{\times }\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{L}}{\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{M}} \\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{L} \end{gathered}$$