Chemistry 10th Edition

by Whitten, Kenneth W.; Davis, Raymond E.; Peck, Larry; Stanley, George G.

Published by Brooks/Cole Publishing Co.

ISBN 10: 1133610668

ISBN 13: 978-1-13361-066-3

Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Conceptual Exercises - Page 112: 95

Answer

$V=425.6ml$

Work Step by Step

$n(Al^{3+})=n(AlCl_3)=\frac{m(AlCl_3)}{M(AlCl_3)}=\frac{0.025\frac{g}{ml}\times 50ml}{133.5\frac{g}{mol}}=9.36\times 10^{-3} mol$ Hence, the final volume of the solution should be: $V=\frac{n}{c}=\frac{9.36\times 10^{-3} mol}{0.022\frac{mol}{dm^3}}=0.4256dm^3=425.6ml$

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