Answer
(a) 136 mL
(b) 62.2 mL
(c) 47 mL
Work Step by Step
(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
