Question

A solution contains 35ml of pentane (density -0.63g/ml) and 65mlof heptane (d=0.68 g/ml).Assuming additive volumes ,find the mass percent ,mole fracttion,mole fraction,molality and molarityof the pentane

Answer 1

Density = Mass / Volume.

a) For pentane:

Volume of pentane = 35.0 mL, Density of pentane = 0.63 g/mL

Density of pentane = Mass of pentane / Volume of pentane.

0.63 g/mL = Mass of pentane / 35.0 mL

Mass of Pentane = 0.63 g/mL * 35.0 mL

Mass of Pentane = 22.05 g

Molar mass of Pentane = 72.15 g/mol.

# of moles of Pentane = Mass of pentane given / Molar mass of Pentane = 22.05 g / 72.15 g/mol = 0.306 mol.

# of moles of Pentane = 0.306 mole.

b) For hexane,

Volume of hexane = 65.0 mL, Density of hexane = 0.68 g/mL

Density of hexane = Mass of hexane / Volume of hexane.

0.68 g/mL = Mass of hexane / 65.0 mL

Mass of hexane = 0.68 g/mL * 65.0 mL

Mass of hexane = 44.2 g

Molar mass of hexane = 86.18 g/mol.

# of moles of hexane = Mass of hexane given / Molar mass of hexane = 44.2 g / 86.18 g/mol = 0.513 mol.

# of moles of Hexane = 0.513 mole.

Now,

1) Mass percent of pentane:

Total mass of solution = Mass of pentane + Mass of hexane = 22.05 g + 44.2 g = 66.25 g = 0.06625 Kg.

%(m/m) Pentane =(Mass of pentane / Total mass of solution) * 100 = (22.05 / 44.2)*100 = 49.89%

Mass of percent of pentane = 49.89% = 50 % (Apprximtely)

2) Mole Fraction of pentane.

Let, # of moles of pentane n1 = 0.306 mole

# of moles of hexane n2 = 0.513 mole

Total # of moles N = n1 + n2 = 0.306+0.513 = 0.819 mol

Mole fraction of pentane = n1 / N = 0.306 / 0.819 = 0.374

Mole fraction of pentane = 0.374

3) Molality (m) of pentane.

Mass of solution = 0.06625 Kg (Calculated above)

# of moles of Pentane = 0.306 mol.

Molality of Pentane = # of moles of pentane / Mass of solution in Kg = 0.306 / 0.06625 = 4.62 molal.

3) Molarity of Pentane.

Volume of the solution = 35.0 mL + 65.0 mL = 100.0 mL = 0.1 L

Molarity of pentane = # of moles of pentane / Volume of the solution L = 0.306 mole / 0.1 L = 3.06 M

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