## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 8 - Properties of Gases - Exercise 8.13 - Partial Pressures - Page 346: a

0.173 g

#### Work Step by Step

$P=626\, mmHg\times\frac{1\,atm}{760\, mmHg}=0.8237\,atm$ $V=355\,mL=0.355 L$ $R=0.0821\,L\,atmmol^{-1}K^{-1}$ $T=(35+273)K=308\,K$ $PV= nRT$ (ideal gas law) $\implies n= \frac{PV}{RT}=\frac{0.8237\,atm\times0.355\,L}{(0.0821\,L\,atm\,mol^{-1}K^{-1})(308\,K)}$ $=0.01156\,mol$ But $n(total)= n(Ne)+n(Ar)$ $n(Ne)=\frac{mass\, of\,Ne}{molar\,mass\,of\,Ne}$ $=\frac{0.146\,g}{20.1797\,g/mol}=0.007235\,mol$ $n(Ar)=n(total)-n(Ne)$ $=0.01156\,mol-0.007235\,mol$ $=0.004325\,mol$ Mass of Ar= $n(Ar)\times molar\,mass\,of\,Ar$ $=0.004325\,mol\times 39.948\,g/mol$ $=0.173\,g$

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