## Chemistry (4th Edition)

(a) $545$ mL = $545 \times 10^{-3}$ L = 0.545 L 1. Calculate the molar mass $(C_2H_5OH)$: 12.01* 2 + 1.008* 5 + 16* 1 + 1.008* 1 = 46.068g/mol 2. Calculate the number of moles $(C_2H_5OH)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 29}{ 46.068}$ $n(moles) = 0.63$ 3. Find the concentration in mol/L $(C_2H_5OH)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.63}{ 0.545}$ $C(mol/L) = 1.16$ (b) $74$ mL = $74 \times 10^{-3}$ L = 0.074 L 4. Calculate the molar mass $(C_{12}H_{22}O_{11})$: 12.01* 12 + 1.008* 22 + 16* 11 = 342.296g/mol 5. Calculate the number of moles $(C_{12}H_{22}O_{11})$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 15.4}{ 342.296}$ $n(moles) = 0.045$ 6. Find the concentration in mol/L $(C_{12}H_{22}O_{11})$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.045}{ 0.074}$ $C(mol/L) = 0.608$ (c) $86.4$ mL = $86.4 \times 10^{-3}$ L = 0.0864 L 7. Calculate the molar mass $(NaCl)$: 22.99* 1 + 35.45* 1 = 58.44g/mol 8. Calculate the number of moles $(NaCl)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 9}{ 58.44}$ $n(moles) = 0.154$ 9. Find the concentration in mol/L $(NaCl)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.154}{ 0.0864}$ $C(mol/L) = 1.78$