## Chemistry (4th Edition)

(a) 1. Determine the molar mass of this compound (NaCl): 22.99* 1 + 35.45* 1 = 58.44g/mol 2. Calculate the number of moles: $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $58.44 = \frac{2.14}{n(mol)}$ $n(mol) = \frac{2.14}{ 58.44}$ $n(mol) = 0.0366$ 3. Find the volume: $Concentration(M) = \frac{n(mol)}{V(L)}$ $0.27 = \frac{0.0366}{V(L)}$ $V(L) = \frac{0.0366}{0.27}$ $V(L) = 0.136$ 4. Convert that number to mL: 1 L = $1000$ mL $0.136$ L = $0.136 \times 1000$ mL = $136$ mL (b) 1. Determine the molar mass of this compound (CH_3CH_2OH): 12.01* 1 + 1.008* 3 + 12.01* 1 + 1.008* 2 + 16* 1 + 1.008* 1 = 46.068g/mol 2. Calculate the number of moles: $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $46.068 = \frac{4.3}{n(mol)}$ $n(mol) = \frac{4.3}{ 46.068}$ $n(mol) = 0.0933$ 3. Find the volume: $Concentration(M) = \frac{n(mol)}{V(L)}$ $1.5 = \frac{0.0933}{V(L)}$ $V(L) = \frac{0.0933}{1.5}$ $V(L) = 0.0622$ 4. Convert that number to mL: 1 L = $1000$ mL $0.0622$ L = $0.0622 \times 1000$ mL = $62.2$ mL (c) 1. Determine the molar mass of this compound (HC_2H_3O_2): 1.008* 1 + 12.01* 2 + 1.008* 3 + 16* 2 = 60.052g/mol 2. Calculate the number of moles: $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $60.052 = \frac{0.85}{n(mol)}$ $n(mol) = \frac{0.85}{ 60.052}$ $n(mol) = 0.014$ 3. Find the volume: $Concentration(M) = \frac{n(mol)}{V(L)}$ $0.3 = \frac{0.014}{V(L)}$ $V(L) = \frac{0.014}{0.3}$ $V(L) = 0.047$ 4. Convert that number to mL: 1 L = $1000$ mL $0.047$ L = $0.047 \times 1000$ mL = $47$ mL