Question

A 5.0 g sample of dry ice (solid carbon dioxide) turned into
2400 cm of carbon dioxide gas at rtp. What was the
percentage purity of the dry ice? (M of CO2 = 44.)​

Answer 1

Answer:

6.14*10-3

Explanation:

Dry ice is solid carbon dioxide. A 0.05 g sample of dry ice is placed in an evacuated 4.6L  vessel at 30  

0

C. Calculate the pressure inside the vessel after all the dry ice has been converted to CO  

2

​  

gas.

answer

PV=nRT

P=  

V

nRT

​  

, we know that R=0.0821 L−atm/k−mol

Given T=30  

o

C⇒30+273=303 k

V=4.6 L

n=  

44.01 g CO  

2

​  

 

1 mol CO  

2

​  

 

​  

=  

44.01

0.059

​  

=0.00113 mol CO  

2

​  

 

∴ P=  

V

nRT

​  

=  

4.6

0.00113×0.0821×303

​  

 

=  

4.6

0.0281

​  

=0.0061087=0.00612

∴  the closest ans is (D) 6.14×10  

−3

atm