## Chemistry: Molecular Approach (4th Edition)

$HClO_4$ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 4 )= 100.46 g/mol a. $$[H_3O^+] = 10^{-pH} = 10^{-2.50} = 3.2 \times 10^{-3} \space M$$ $$0.500 \space L \times \frac{3.2 \times 10^{-3} \space mol}{1 \space L} \times \frac{100.46 \space g}{1 \space mol} = 0.16 \space g$$ b. $$[H_3O^+] = 10^{-pH} = 10^{-1.50} = 3.2 \times 10^{-2} \space M$$ $$0.500 \space L \times \frac{3.2 \times 10^{-2} \space mol}{1 \space L} \times \frac{100.46 \space g}{1 \space mol} = 1.6 \space g$$ c. $$[H_3O^+] = 10^{-pH} = 10^{-0.50} = 0.32 \space M$$ $$0.500 \space L \times \frac{0.32 \space mol}{1 \space L} \times \frac{100.46 \space g}{1 \space mol} = 16 \space g$$