## Chemistry: Molecular Approach (4th Edition)

a. 0.062 M $LiNO_3$ b. 0.675 M $C_2H_6O$ c. $6.898\times 10^{- 4}M$ $KI$
a. $C(mol/L) = \frac{n(moles)}{volume (L)} = \frac{0.38 moles}{6.14L} = 0.062M$ ---- b. 1. Calculate the molar mass $(C_2H_6O)$: 12.01* 2 + 1.008* 6 + 16* 1 = 46.07g/mol 2. Calculate the number of moles $(C_2H_6O)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 72.8}{ 46.07}$ $n(moles) = 1.58$ 3. Find the concentration in mol/L $(C_2H_6O)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 1.58}{ 2.34}$ $C(mol/L) = 0.675$ ----------------------- c. $112.4$ mL = $112.4 \times 10^{-3}$ L = 0.1124 L 1. Calculate the molar mass $(KI)$: 39.1* 1 + 126.9* 1 = 166.0g/mol 2. Calculate the number of moles $(KI)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.01287}{ 166.0}$ $n(moles) = 7.753\times 10^{- 5}$ 3. Find the concentration in mol/L $(KI)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 7.753\times 10^{- 5}}{ 0.1124}$ $C(mol/L) = 6.898\times 10^{- 4}M$