Q13.58 P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the molarity of each aqueous solution: 

(a) 78.0 mL of 0.240 M NaOH diluted to 0.250 L with water.

(b) 38.5 mL of 1.2 M ${\mathbf{HNO}}_{\mathbf{3}}$  diluted to 0.130 L with water.

Step-by-Step Solution

Answer

1. The molarity of the NaOH is 0.240M in 0.078L of water when the solution is diluted to 0.250L the molarity of NaOH is 0.075M.
2. The molarity of the $HN{O}_3$  is 1.2M in 0.0385L of water when the solution is diluted to 0.130L the molarity of  $HN{O}_3$ is 0.355M.

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.

 

On dilution, the number of moles of solute remains the same. 

The number of moles can be obtained by multiplying molarity of the solution with the volume of the solution in litres.

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

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

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

Step 2: Subpart (a)

Volume of solute  $\left(\mathbf{NaOH}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{078}\mathbf{L}$

Molarity of NaOH at  $\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{L}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{240}\mathbf{M}$

Volume of Solution is 0.250 L

Let the molarity of solute at 0.250 L volume be  $M_2$

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

$$\begin{gathered} \mathbf{0}\mathbf{.}\mathbf{240}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{078}\mathbf{L}\mathbf{=}\mathbf{M}\mathbf{2}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{L} \\ \mathbf{M}\mathbf{2}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{240}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{078}\mathbf{L}}{\mathbf{0}.2\mathbf{50}\mathbf{L}} \\ \mathbf{M}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{075}\mathbf{M} \end{gathered}$$

Volume of Solution  $\mathbf{38}\mathbf{.}\mathbf{5}\mathbf{mL}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0385}\mathbf{L}$

 

Molarity of  $HN{O}_3$  at  $\mathbf{0}\mathbf{.}\mathbf{078}\mathbf{L}\mathbf{ } \mathbf{solution}\mathbf{ } \mathbf{is}\mathbf{ } \mathbf{1}\mathbf{.}\mathbf{2}\mathbf{M}$

Volume of Solution is 0.130 L.

Step 3: Subpart (b)

Molarity of solute  $HN{O}_3$ at  $\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{L}\mathbf{=}{\mathbf{V}}_{\mathbf{2}}$ be  $M_2$.

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

$$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{2}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0385}\mathbf{L}\mathbf{=}\mathbf{M}\mathbf{2}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{130}\mathbf{L} \\ \mathbf{M}\mathbf{2}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{M}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0385}\mathbf{L}}{\mathbf{0}\mathbf{.}\mathbf{130}\mathbf{L}} \\ \mathbf{M}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{355}\mathbf{M} \end{gathered}$$