## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 1 - Exercises - Page 41: 133

#### Answer

$\underline{0.492}$

#### Work Step by Step

$\text{1 L}=1000\text{ mL}$ The volume of 175 L is converted into milliliters as follows: $\left( 175\text{ L} \right)\left( \frac{1000\text{ mL}}{1\text{ L}} \right)=175,000\text{ mL}$ The mass of nitrogen is calculated as follows: $m=d\,\,\times \,\,V$ Thus, mass is calculated as follows: \begin{align} & m=\left( 0.808\text{ g/mL} \right)\left( 175,000\text{ mL} \right) \\ & =141,400\text{ g} \end{align} Now, volume of gaseous nitrogen is calculated as follows: \begin{align} & V=\frac{141,400}{1.15\text{ g/L}} \\ & =122,956.52\text{ L} \end{align} $\text{1 }{{\text{m}}^{3}}=1000\text{ L}$ Thus, volume in cubic millimeters is written as follows: \begin{align} & V=\left( 122,956.52\text{ L} \right)\left( \frac{1\text{ }{{\text{m}}^{3}}}{1000\text{ L}} \right) \\ & =122.95652\text{ }{{\text{m}}^{3}} \end{align} The volume of the room is calculated with the help of the dimensions of the room as follows: $10.00\text{ m}\times \text{10}\text{.00 m}\times \text{2}\text{.50 m}=250\text{ }{{\text{m}}^{3}}$ Thus, the fraction of air in the room that is displaced by gaseous nitrogen is equal to the ratio of volume of gaseous nitrogen to the volume of the room as follows: \begin{align} & F=\frac{122.95652\text{ }{{\text{m}}^{3}}}{250\text{ }{{\text{m}}^{3}}} \\ & =0.492 \end{align} The fraction of air in the room that is displaced by the gaseous nitrogen is $\underline{0.492}$.

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