Q13.70 P - Chemistry: Molecular Nature Of Matter And Change

Question

A solution contains 0.35 mol of isopropanol (${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}$ ) dissolved in 0.85 mol of water.

(a) What is the mole fraction of iso-propanol? 

(b) The mass percent? 

(c) The molality?

Step-by-Step Solution

Answer

1. The mole-fraction of 0.35mol of iso-propanol present in 0.85mol water is 0.3.
2. The mass percentage of the 21g of iso-propanol in 15.3g is 58%.
3. The molality of the 0.35mol of iso-propanol in 0.0153kg is 22.9m.

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.

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

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

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

Mole Fraction of a solute can be defined as the ratio of the number of moles of the solute to the number of moles of solution.

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

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.

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

Step 2: Expression to calculate the Concentration

1. Number of moles of  ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{moles}$

Number of moles of water  $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{85}\mathbf{moles}$

$$\begin{gathered} \mathbf{Mole} \mathbf{Fraction}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{ } \mathbf{of} \mathbf{One} \mathbf{Component}}{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of}\mathbf{ } \mathbf{the}\mathbf{ } \mathbf{Solvent}\mathbf{+}\mathbf{Number} \mathbf{of}\mathbf{ } \mathbf{Moles}\mathbf{ } \mathbf{of} \mathbf{Solute}} \\ \mathbf{Mole}\mathbf{ } \mathbf{Fraction} \mathbf{of}\mathbf{ }\mathbf{ } {\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{35} \mathbf{mol}}{\left(\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{85}\right)\mathbf{mol}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{35}}{\mathbf{1}\mathbf{.}\mathbf{2}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3} \end{gathered}$$

Hence, the mole of fraction  ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}$ is 0.3

b) Number of moles of isopropanol is 0.35 mol.

$$\begin{gathered} \mathbf{Number}\mathbf{ } \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ {\mathbf{n}}_{{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{60}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{0}\mathbf{.}\mathbf{35}\mathbf{mol}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{60}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{Mass} \mathbf{of} {\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{mol}\mathbf{\times }\mathbf{60}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{Mass} \mathbf{of} {\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}\mathbf{=}\mathbf{21}\mathbf{g} \end{gathered}$$

Mass of  ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}$, Solute  $\mathbf{=}\mathbf{21}\mathbf{g}$

Number of moles of water  $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{85}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}} \\ \mathbf{NumberofMoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{0}\mathbf{.}\mathbf{85}\mathbf{mol}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{85}\mathbf{mol}\mathbf{\times }\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{3}\mathbf{g} \end{gathered}$$

Therefore, Mass of water, Solvent $\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{3}\mathbf{g}$

$$\begin{gathered} \mathbf{Mass}\%\mathbf{=}\frac{\mathbf{Mass} \mathbf{of} \mathbf{Solute}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solution}}\mathbf{\times }\mathbf{100} \\ \mathbf{=}\frac{\mathbf{21}\mathbf{g}}{\mathbf{21}\mathbf{g}\mathbf{+}\mathbf{15}\mathbf{.}\mathbf{3}\mathbf{g}}\mathbf{\times }\mathbf{100} \\ \mathbf{=}\frac{\mathbf{21}\mathbf{g}}{\mathbf{36}\mathbf{.}\mathbf{3}\mathbf{g}}\mathbf{\times }\mathbf{100} \\ \mathbf{=}\mathbf{58}\mathbf{\%} \end{gathered}$$

c) Moles of Solute,  ${\mathbf{n}}_{{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{OH}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{mol}$

Mass of Solvent, Water  ${\mathbf{Mass}}_{\mathbf{water}}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{3}\mathbf{g}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0153}\mathbf{kg}$

$$\begin{gathered} \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{solute}}{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{Solvent}\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{kg}\mathbf{ }\mathbf{)}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{0153}\mathbf{kg}} \\ \mathbf{=}\mathbf{22}\mathbf{.}\mathbf{9}\mathbf{m} \end{gathered}$$

Hence, the Molality is 22.9m