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 - Building Your Knowledge - Page 114: 106

Answer

$c(NaCl)=0.5230M$

Work Step by Step

$n(NaCl)=n_1+n_2=c_1V_1+c_2V_2=0.375\frac{mol}{dm^3}\times 0.035dm^3 +0.632\frac{mol}{dm^3}\times 0.0475dm^3 =0.043145mol$ $V=V_1+V_2=0.035dm^3+0.0475dm^3=0.0825dm^3$ Hence, $c(NaCl)=\frac{n(NaCl)}{V}=\frac{0.043145mol}{0.0825dm^3}=0.5230M$

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