Question

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.

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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