Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 1 Basic Concepts of Chemistry - Study Questions: 46

Answer

A) $\rho=2.331\ g/mL$ B) 11.73 mL of $CHCl_3$ and 8.27 mL of $CHBr_3$

Work Step by Step

Density definition: $\rho=\frac{m}{V}$ A) Calculating the masses from the given data: $CHCl_3$ ($\rho=1.492\ g/mL,\ V=10.0\ mL$):$m=14.92\ g$ $CHBr_3$ ($\rho=2.890\ g/mL,\ V=15.0\ mL$): $m=43.35\ g$ Calculating the density of the mixture $m=58.27\ g,\ V=25.0\ mL$: $\rho=2.331\ g/mL$ B) Total volume: 20mL Defining a variable for one's volume and calculating the volume of the other as a function of that variable: $CHCl_3$ volume: V $CHBr_3$ volume: 20-V Calculating the total mass of the mixture from the given data: ($\rho=2.07\ g/mL,\ V=20.0\ mL$): $m=41.4\ g$ Applying that the mass of the mixture must be equal to the sum of the individual masses of its components: $41.4=1.492\times V + 2.890\times (20-V)$ And Silvinha for V: $V=11.73\ mL$
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