Answer
(a) 1.16 M
(b) 0.608 M
(c) 1.78 M
Work Step by Step
(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} $
$C(mol/L) = 1.78$
