Answer
1.588g of $HClO_4$
Work Step by Step
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.5}$
$[H_3O^+] = 3.162 \times 10^{- 2}M$
2. Since $HClO_4$ is a strong acid: $[HClO_4] = [H_3O^+] = 3.162 \times 10^{- 2}M$
3. Calculate the number of moles:
$n(moles) = concentration(M) * volume(L)$
$n(moles) = 0.03162 * 0.5$
$n(moles) = 0.01581$
4. Find the mass value in grams:
Molar mass: 1.01* 1 + 35.45* 1 + 16* 4 = 100.46g/mol
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 100.46 * 0.01581$
$mass(g) = 1.588$
