Q13.74 P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the molality, molarity, and mole fraction of ${\mathbf{NH}}_{\mathbf{3}}$ in an 8.00 mass % aqueous solution (d = 0.9651 g/mL)?

Step-by-Step Solution

Answer

The number of moles of ${\mathbf{NH}}_{\mathbf{3}}$ and water is 0.471mole and 5.1mole. The molality and mole fraction and molarity of ${\mathbf{NH}}_{\mathbf{3}}$ is 5.12m, 0.085 and 4.55M respectively.

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.

Molality can be defined as the ratio of the mass of the solute to the mass of the solution present in kilogram measure.

$\mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{kilogram} \mathbf{)}}$

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

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

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

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

Mole Fraction can be defined as the ratio of the number of moles of the solute to the number of moles of solvent and number of solute.

$\mathbf{Mole} \mathbf{Fraction}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{One} \mathbf{Component}}{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{the} \mathbf{Solvent}\mathbf{+}\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{Solute}}$

Parts by mass: It can be defined as the percentage of the ratio of the mass of solute to the total mass of the solution.

$\mathbf{Parts} \mathbf{by} \mathbf{Mass}\mathbf{=}\frac{\mathbf{Mass} \mathbf{of} \mathbf{Solute}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solution}}\mathbf{\times }\mathbf{100}$

Density can be defined as the ratio of the mass of the matter to the volume of the matter.

$\mathbf{Density}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}}$

Step 2: Expression to calculate the Concentration

8% mass of  ${\mathbf{NH}}_{\mathbf{3}}$ means that 8g of ${\mathbf{NH}}_{\mathbf{3}}$  present in the 100g of solution.

Now, 

$$\begin{gathered} \mathbf{Massofsolution}\mathbf{=}\mathbf{Massofsolvent}\mathbf{,}\mathbf{water}\mathbf{+}\mathbf{Massofsolute} \\ \mathbf{MassofSolvent}\mathbf{,}\mathbf{water}\mathbf{=}\mathbf{Massofsolution}\mathbf{-}\mathbf{Massofsolute}\mathbf{,}{\mathbf{NH}}_{\mathbf{3}} \\ \mathbf{MassofSolvent}\mathbf{,}\mathbf{water}\mathbf{=}\mathbf{100}\mathbf{g}\mathbf{-}\mathbf{8}\mathbf{g} \\ \mathbf{MassofSolvent}\mathbf{,}\mathbf{water}\mathbf{=}\mathbf{92}\mathbf{g} \end{gathered}$$

Mass of Solute,  ${\mathbf{mass}}_{{\mathbf{NH}}_{\mathbf{3}}}\mathbf{=}\mathbf{8}\mathbf{g}$

Molar Mass of Solute,  ${\mathbf{M}}_{{\mathbf{NH}}_{\mathbf{3}}}\mathbf{=}\mathbf{17}\mathbf{g}\mathbf{/}\mathbf{mol}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{NumberofMoles} \mathbf{of} {\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{8}\mathbf{g}}{\mathbf{17}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{NumberofMoles} \mathbf{of} {\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{471}\mathbf{mol} \end{gathered}$$

Mass of Solvent, Water ${\mathbf{Mass}}_{\mathbf{solvent}}\mathbf{=}\mathbf{92}\mathbf{g}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{092}\mathbf{kg}$

$$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}} \\ \mathbf{NumberofMoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\frac{\mathbf{92}\mathbf{g}}{\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{NumberofMoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{1}\mathbf{mol} \end{gathered}$$

Number of moles of solute,  ${\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{471}\mathbf{mol}$

Number of moles of solvent, Water  ${\mathbf{n}}_{\mathbf{water}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{1} \mathbf{mol}$

$$\begin{gathered} \mathbf{Mole}\mathbf{ }\mathbf{Fraction}\mathbf{=}\frac{\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{One}\mathbf{ }\mathbf{Component}}{\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{Solvent}\mathbf{+}\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Solute}} \\ \mathbf{Mole}\mathbf{ }\mathbf{Fraction}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{NH}\mathbf{3}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{471}}{\mathbf{5}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{471}} \\ \mathbf{Mole}\mathbf{ }\mathbf{Fraction}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{NH}\mathbf{3}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{471}}{\mathbf{5}\mathbf{.}\mathbf{571}} \\ \mathbf{Mole}\mathbf{ }\mathbf{Fraction}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{NH}\mathbf{3}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{085} \end{gathered}$$

Number of moles of  ${\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{471}\mathbf{mol}$

$$\begin{gathered} \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{kilogram} \mathbf{)}} \\ \mathbf{Molality}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{471}\mathbf{mole}}{\mathbf{0}\mathbf{.}\mathbf{092}\mathbf{kg}} \\ \mathbf{Molality}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{12}\mathbf{m} \end{gathered}$$

Density of the Solution  $\mathbf{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9651}\mathbf{g}\mathbf{/}\mathbf{mL}$

$$\begin{gathered} \mathbf{Density} \mathbf{of} \mathbf{Solution}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}} \\ \mathbf{0}\mathbf{.}\mathbf{9651}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{=}\frac{\mathbf{100}\mathbf{g}}{\mathbf{Volume}} \\ \mathbf{Volume}\mathbf{=}\frac{\mathbf{100}\mathbf{g}}{\mathbf{0}\mathbf{.}\mathbf{9651}\mathbf{g}\mathbf{/}\mathbf{mL}} \\ \mathbf{Volume}\mathbf{=}\mathbf{103}\mathbf{.}\mathbf{62}\mathbf{mL} \end{gathered}$$

Volume of solution  $\mathbf{Volume} \mathbf{of} \mathbf{solution}\mathbf{=}\mathbf{103}\mathbf{.}\mathbf{62}\mathbf{mL}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10362}\mathbf{L}$

$$\begin{gathered} \mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{solute}}{\mathbf{Volume} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{471}\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{10362}\mathbf{L}} \\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{55}\mathbf{M} \end{gathered}$$

The molarity of  ${\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{55}\mathbf{M}$