## Chemistry: Molecular Approach (4th Edition)

a. 1.17 M $LiCl$ b. 0.123 M $C_6H_{12}O_6$ c. $4.53 \times 10^{-3}M = 0.00453M$ $NaCl$
a. $C(mol/L) = \frac{n(moles)}{volume (L)} = \frac{3.25 moles}{2.78L} = 1.17$ -------------------- b. 1. Calculate the molar mass $(C_6H_{12}O_6)$: 12.01* 6 + 1.008* 12 + 16.00* 6 = 180.16g/mol 2. Calculate the number of moles $(C_6H_{12}O_6)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 28.33}{ 180.16}$ $n(moles) = 0.1572$ 3. Find the concentration in mol/L $(C_6H_{12}O_6)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.1572}{ 1.28}$ $C(mol/L) = 0.123$ --------------------------------------------- c. $122.4$ mL = $122.4 \times 10^{-3}$ L = 0.1224 L 1. Calculate the molar mass $(NaCl)$: 22.99* 1 + 35.45* 1 = 58.44g/mol 2. Calculate the number of moles $(NaCl)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.0324}{ 58.44}$ $n(moles) = 5.544\times 10^{- 4}$ 3. Find the concentration in mol/L $(NaCl)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 5.544\times 10^{- 4}}{ 0.1224}$ $C(mol/L) = 4.53\times 10^{- 3}$