Answer
(a)
$$ 1.76 \space M \space NaCl$$
(b)
$$ 0.770 \space M \space H_2SO_4$$
(c) $$1.34 \times 10^{-3} \space M \space C_6H_5OH $$
Work Step by Step
(a) $ NaCl $ : ( 22.99 $\times$ 1 )+ ( 35.45 $\times$ 1 )= 58.44 g/mol
- Calculate the amount of moles:
$$ 25.65 \space g \times \frac{1 \space mol}{ 58.44 \space g} = 0.4389 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.4389 \space mol}{ 250 \space mL} \times \frac{1000 \space mL}{ 1 \space L}= 1.76 \space M $$
(b) $ H_2SO_4 $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 4 )+ ( 32.07 $\times$ 1 )= 98.09 g/mol
- Calculate the amount of moles:
$$ 75.5 \space g \times \frac{1 \space mol}{ 98.09 \space g} = 0.770 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.770 \space mol}{ 1.00 \space L} = 0.770 \space M $$
(c) $ C_6H_5OH $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 1 )= 94.11 g/mol
- Calculate the amount of moles:
$$ 0.126 \space g \times \frac{1 \space mol}{ 94.11 \space g} = 1.34 \times 10^{-3} \space mol$$
- Calculate the molarity:
$$ \frac{ 1.34 \times 10^{-3} \space mol}{ 1.00 \space L} = 1.34 \times 10^{-3} \space M $$
