General Chemistry: Principles and Modern Applications (10th Edition)

(a) 1. Molarity, by definition: 0.112 M $C_6H_{12}O_6$: 0.112 mol $C_6H_{12}O_6$ for 1 L of solution. 2. Find the mass of $C_6H_{12}O_6$ $C_6H_{12}O_6$ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.112 \space mole \times \frac{ 180.16 \space g}{1 \space mole} = 20.2 \space g$$ 3. Calculate the mass of solution, and the mass of water. 1 L of solution: $$1 \space L \space solution \times \frac{1.006 \space g}{1 \space mL} \times \frac{1000 \space mL}{1 \space L} = 1006 \space g$$ Mass of water = 1006 g - 20.2 g = 986 g 4. Find the amount of moles of water. $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$986 \space g \times \frac{1 \space mole}{ 18.02 \space g} = 54.7 \space moles$$ 5. Calculate the mole fraction. $$Mole \space fraction \space solute = \frac{0.112 \space mol}{0.112 \space mol + 54.7 \space mol} = 0.00204$$ (b) 1. By definition: 3.20% ethanol by volume: 3.20 mL of ethanol for each 100 mL of solution. 2. Calculate the amount of moles of ethanol $CH_3CH_2OH$ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol $$3.20 \space mL \times \frac{0.789 \space g}{1 \space mL} \times \frac{1 \space mol}{46.07 \space g} = 0.0548 \space mol$$ 3. Calculate the amount of moles of water. Mass of solution: $$100 \space mL \times \frac{0.993 \space g}{1 \space mL} = 99.3 \space g$$ Mass of ethanol: $$3.20 \space mL \times \frac{0.789 \space g}{1 \space mL} = 2.52 \space g$$ Mass of water = 99.3 g - 2.52 g = 96.8 g $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol $$96.8 \space g \times \frac{1 \space mol}{18.02 \space g} = 5.37 \space mol$$ 4. Calculate the mole fraction: $$Mole \space fraction \space ethanol = \frac{0.0548 \space mol}{0.0548 \space mol + 5.37 \space mol} = 0.0101$$