Question

Calculate the mole fraction of solute in the following aqueous solutions (a) 0.112 M C₆H₁₂O₆ (glucose solution, p = 1.006 g/mL; (b) 3.20% ethanol (CH₃CH₂OH) by volume (p = 0.993 g/mL; pure ethanol, p = 0.789 g/mL)

Answer 1

a) let us take 1L solution ,

mass = vol x density = 1000 x 1.006 g/ml = 1006 g

Moles of glucose = M x V = 0.112 x 1 = 0.112 ,

mass of glucose = moles x molar mas s= 0.112 x 180.16 = 20.178 g

mass of water = 1006-20.178 = 985.822 g , H2O moles = 985.822/18 = 54.768

mol fraction = ( 0.112 /0.112+54.768) = 0.002

b) 3.2 ml ethanol per 100 ml solution

mass of ethanol = 3.2 x 0.789 = 2.525 g , ethanol moles = mass/molar mass = 2.525/46.07 = 0.0548

mass of solution = vol x dnesity = 100 x 0.993 = 99.3 g

water mass = 99.3-2.525 =96.775 g , water moles = 96.775/18 = 5.3764

mole fraction = ( 0.0548/0.0548+5.3764) = 0.01