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.
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
Get Answers For Free
Most questions answered within 1 hours.