Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 15 - Sections 15.1-15.12 - Exercises - Problems by Topic - Page 747: 60c

Answer

15.88g of $HClO_4$

Work Step by Step

1.Calculate the hydronium concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 0.5}$ $[H_3O^+] = 0.3162M$ 2. Since $HClO_4$ is a strong acid: $[HClO_4] = [H_3O^+] = 0.3162M$ 3. Calculate the number of moles: $n(moles) = concentration(M) * volume(L)$ $n(moles) = 0.3162 * 0.5$ $n(moles) = 0.1581$ 4. Find the mass value in grams: 1.01* 1 + 35.45* 1 + 16* 4 ) = 100.46g/mol $mass(g) = mm(g/mol) * n(moles)$ $mass(g) = 100.46 * 0.1581$ $mass(g) = 15.88$
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