Answer
(a) $$ 0.0108 \space M Al_2(SO_{4})_3$$
(b)
$$ 0.0418 \space M CaCl_2 .6H_{2}O$$
(c)
$$0.336 \space M HBr$$
Work Step by Step
(a) $ Al_2(SO_{4})_3 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 4 $\times$ 3 )+ ( 32.07 $\times$ 3 )= 342.17 g/mol
- Calculate the amount of moles:
$$ 1.57 \space g \times \frac{1 \space mol}{ 342.17 \space g} = 4.59 \times 10^{-3} \space mol$$
* 425 mL = 0.425 L
- Calculate the molarity:
$$ \frac{ 4.59 \times 10^{-3} \space mol}{ 0.425 \space L} = 0.0108 \space M $$
(b) $ CaCl_2 .6H_{2}O $ : ( 35.45 $\times$ 2 )+ ( 40.08 $\times$ 1 )+ ( 1.008 $\times$ 2 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 219.08 g/mol
- Calculate the amount of moles:
$$ 45.8 \space g \times \frac{1 \space mol}{ 219.08 \space g} = 0.209 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.209 \space mol}{ 5.00 \space L} = 0.0418 \space M $$
(c)$ HBr $ : ( 79.90 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 80.91 g/mol
- Calculate the amount of moles:
$$ 18.4 \space g \times \frac{1 \space mol}{ 80.91 \space g} = 0.227 \space mol$$
* 675 mL = 0.675 L
- Calculate the molarity:
$$ \frac{ 0.227 \space mol}{ 0.675 \space L} = 0.336 \space M $$
