Question

Calculate the mole fraction of solute in the following aqueous
solutions (a) 0.112 M C_{6}H_{12}O_{6}
(glucose solution, p = 1.006 g/mL; (b) 3.20% ethanol
(CH_{3}CH_{2}OH) 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**

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