Question

A solution is prepared by mixing 25 mL pentane (C5H12, d = 0.63 g/cm3) with 30...

A solution is prepared by mixing 25 mL pentane (C5H12, d = 0.63 g/cm3) with 30 mL hexane (C6H14, d = 0.66 g/cm3). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the pentane.

Homework Answers

Answer #1

we know that

1 cm3 = 1 ml

also

mass = density x volume

so

mass of pentane = 25 x 0.63 = 15.75 g

mass of hexane = 30 x 0.66 = 19.8 g

now

total mass = 15.75 + 19.8

total mass = 35.55

now

% mass = mass of pentane x 100 / total mass

% mass = 15.75 x 100 / 35.55

% mass = 44.3

so

the mass percent of pentane is 44.3 %

now

moles = mass / molar mass

so

moles of pentane = 15.75 / 72 = 0.21875

moles of hexane = 19.8 / 86 = 0.23023

now

total moles = 0.21875 + 0.23023 = 0.44898

now

mol fraction of pentane = moles of pentane / total moles

mol fraction of pentane = 0.21875 / 0.44898

mol fraction of pentane = 0.487


now

molality = moles of pentane x 1000 / mass of hexane (g)

molality = 0.21875 x 1000 / 19.8

molallity of pentane = 11.048

now

total volume = 25 + 30 = 55 ml

now

molarity = moles of pentane x 1000 / volume of sollution (ml)

molarity = 0.21875 x 1000 / 55

molarity of pentane = 3.98

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