Answer:
[tex]\%purity=86.4\%[/tex]
Explanation:
Hello,
In this case, since the dry ice is the solid phase of carbon dioxide gas, via the yielded volume we can compute the grams that were sublimated via the ideal gas equation at RTP (1 atm and 25 °C) considering the volume in liters (2.1 L):
[tex]PV=nRT\\\\n=\frac{PV}{RT}\\ \\\frac{m}{M} =\frac{PV}{RT}\\\\m=\frac{MPV}{RT}\\\\m=\frac{44g/mol*1atm*2.4L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\m=4.32g[/tex]
Those previously computed mass are the pure grams, therefore, the percentage purity is:
[tex]\% purity=\frac{m_{pure}}{m_{total}}*100\%\\ \\\%purity=\frac{4.32g}{5.0g}*100\%\\ \\\%purity=86.4\%[/tex]
Best regards.
