Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 17 - Questions and Problems - Page 825: 17.27


The molar mass of this acid is equal to 202g/mol.

Work Step by Step

1. Calculate the number of moles of $KOH$: 16.4ml = 0.0164L $C_1 * V_1 = n(moles)$ $ 0.08133* 0.0164= n(moles)$ $ 0.001334 = n(moles)$ 2. Since the acid is monoprotic, to neutralize it, the number of moles of the $KOH$ should be equal. - n(moles of acid)= $1.334 \times 10^{-3}$ 3. Calculate the molar mass: $MM= \frac{mass(g)}{n(moles)} = \frac{0.2688}{1.334 \times 10^{-3}} =\\ 201.5g/mol$
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