Q13.57 - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the molarity of each aqueous solution:

1.   0.82 g of ethanol (${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}$ ) in 10.5 mL of solution
2. 1.27 g of gaseous  ${\mathbf{NH}}_{\mathbf{3}}$ in 33.5 mL of solution.

Step-by-Step Solution

Answer

1. The molarity of the 0.82g of table sugar $\left({\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\right)$  in 0.0105L of the solution is 1.7M.

2. The molarity of the 1.27g of ${\mathbf{NH}}_{\mathbf{3}}$  in 0.0335L of the solution is 2.23M.

Step 1: Concentration of the Solution

The 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 the litre measure.

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

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

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

Step 2: Expression to calculate the Concentration

a) Mass of  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{82}\mathbf{g}$

Molar Mass of  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{=}\mathbf{46}\mathbf{g}\mathbf{/}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{NumberofMoles}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{82}\mathbf{g}}{\mathbf{46}\mathbf{g}\mathbf{/}\mathbf{mole}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0178}\mathbf{mole} \end{gathered}$$

$$\begin{gathered} \mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Solvent}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{5}\mathbf{mL} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0105}\mathbf{L} \end{gathered}$$

$$\begin{gathered} \mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Volume} \mathbf{of} \mathbf{the} \mathbf{Solution}\mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{Molarity}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{0178}\mathbf{mole}}{\mathbf{0}\mathbf{.}\mathbf{0105}} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{M} \end{gathered}$$

b) Mass of  ${\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{g}$

Molar Mass of  ${\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{17}\mathbf{g}\mathbf{/}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number}\mathbf{ } \mathbf{of}\mathbf{ } \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ } \mathbf{Mass}} \\ \mathbf{NumberofMoles}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{g}}{\mathbf{17}\mathbf{g}\mathbf{/}\mathbf{mole}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0747}\mathbf{mole} \end{gathered}$$

$$\begin{gathered} \mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Solvent}\mathbf{=}\mathbf{33}\mathbf{.}\mathbf{5}\mathbf{mL} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0335}\mathbf{L} \end{gathered}$$

$$\begin{gathered} \mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number} \mathbf{of}\mathbf{ } \mathbf{Moles}}{\mathbf{Volume} \mathbf{of} \mathbf{the}\mathbf{ } \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{Molarity}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{0747}\mathbf{mole}}{\mathbf{0}\mathbf{.}\mathbf{0335}} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{23}\mathbf{M} \end{gathered}$$