## General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

a. 1600 g of $Al(NO_3)_3$ b. 6.8 g $C_6H_{12}O_6$ c. 17.9 g of $LiCl$
a. 1. Find the conversion factors - 3.0 M $Al(NO_3)_3$ solution: $$3.0 \space moles \space Al(NO_3)_3 = 1 \space L \space solution$$ - Molar mass: $Al(NO_3)_3$ : ( 26.98 $\times$ 1 )+ ( 14.01 $\times$ 3 )+ ( 16.00 $\times 3 \times 3$ )= 213.01 g/mol 2. Calculate the mass in grams: $$2.5 \space L \space solution \times \frac{3.0 \space moles \space Al(NO_3)_3}{1 \space L \space solution} \times \frac{213.01 \space g \space Al(NO_3)_3}{1 \space mole \space Al(NO_3)_3} = 1600 \space g \space Al(NO_3)_3$$ ------ b. 1. Find the conversion factors - 0.50 M $C_6H_{12}O_6$ solution: $$0.50 \space mole \space C_6H_{12}O_6 = 1 \space L \space solution$$ - Molar mass: $C_6H_{12}O_6$ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol - 1 L = 1000 mL 2. Calculate the mass in grams: $$75 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{0.50 \space mole \space C_6H_{12}O_6}{1 \space L \space solution} \times \frac{180.16 \space g \space C_6H_{12}O_6}{1 \space mole \space C_6H_{12}O_6} = 6.8 \space g \space C_6H_{12}O_6$$ ------------- c. 1. Find the conversion factors - 1.80 M $LiCl$ solution: $$1.80 \space moles \space LiCl = 1 \space L \space solution$$ - Molar mass: $LiCl$ : ( 6.941 $\times$ 1 )+ ( 35.45 $\times$ 1 )= 42.39 g/mol - 1 L = 1000 mL 2. Calculate the mass in grams: $$235 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{1.80 \space moles \space LiCl}{1 \space L \space solution} \times \frac{42.39 \space g \space LiCl}{1 \space mole \space LiCl} = 17.9 \space g \space LiCl$$