Question

A solution is prepared by mixing 25 mL pentane (C₅H₁₂, d = 0.63 g/cm³) with 45 mL hexane (C₆H₁₄, =0.66 g/cm³). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the pentane.

Short answer

Mass percent of pentane= 34.65%

Mole fraction of pentane= 0.386

Molality of pentane =7.306 m

Molarity of pentane = 3.1 M

Step-by-step solution

Step 1:

To find the mass percent of pentane

$$\begin{gathered} \mathrm{Molality}=\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kilograms}} \\ \mathrm{Given}\mathrm{density}\mathrm{}\mathrm{of}\mathrm{}\mathrm{pentane}\mathrm{}=0.63{\text{g/cm}}^3 \\ \mathrm{Volume}\mathrm{of}\mathrm{pentane}=25\text{ml} \\ \mathrm{grams}\mathrm{of}\mathrm{penane}=0.63\times 25=15.75\text{grams} \\ \mathrm{Similarly}\mathrm{grams}\mathrm{of}\mathrm{hexane}=0.66\times 45=29.7\text{g} \\ \mathrm{Grams}\mathrm{of}\mathrm{solution}=15.75+29.7=45.5\text{g} \\ \mathrm{Therefore}\mathrm{mass}\mathrm{percentage}\mathrm{of}\mathrm{pentane}=\frac{15.75}{45.5}\times 100= 34.65\% \end{gathered}$$

:

To find the mole fraction of pentane

$$\begin{gathered} \mathrm{Mole}\mathrm{}\mathrm{fraction}\mathrm{}\mathrm{of}\mathrm{}\mathrm{component}=\frac{na}{na+nb} \\ \mathrm{Moles}\mathrm{of}\mathrm{pentane}=\frac{\mathrm{Given}\mathrm{weight}}{\mathrm{Molar}\mathrm{mass}} \\ =\frac{15.75}{72.15} \\ =0.217 \\ \mathrm{Moles}\mathrm{of}\mathrm{hexane}=86.18=0.344\text{mol} \\ \mathrm{Mole}\mathrm{fraction}\mathrm{of}\mathrm{pentane}=\frac{0.217}{0.217+0.344} \end{gathered}$$

:

To find the molality of pentane

$$\begin{gathered} \mathrm{Molality}=\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kilograms}} \\ \mathrm{Moles}\mathrm{of}\mathrm{pentane}=0.217\text{mol} \\ \mathrm{Weight}\mathrm{}\mathrm{of}\mathrm{}\mathrm{hexane}\mathrm{}\mathrm{in}\text{kg}=\frac{29.7}{1000} \\ \mathrm{Molality}=\frac{0.217}{0.0297} \\ =7.306\text{m} \end{gathered}$$

:

To find the molarity of pentane

$$\begin{gathered} \mathrm{Molality}=\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kilograms}} \\ \mathrm{Moles}\mathrm{of}\mathrm{pentane}=0.217\text{mol} \\ \mathrm{Weight}\mathrm{of}\mathrm{hexane}\mathrm{in}\mathrm{kg}=\frac{25+45}{1000}=0.07\text{L} \\ \mathrm{Molality}=\frac{0.217}{0.07} \\ =3.1\text{m} \end{gathered}$$