A mixture of CO and CO2 is found to have a density of 1.5 gm/L at 30oC and 740 mm pressure. The percentage composition of CO2 in the mixture is:
A
64.5%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
61.5%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
74.5%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
71.5%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 64.5% PV=nRT ⟹PV=massmoleularweightRT ⟹massV=P×moleularweightRT ⟹d=P×MavgRT, where Mavg is the molecular weight of the mixture. ⟹1.5=740760×Mavg0.0821×303 Let x be the percentage composition of CO2 in the mixture. ⟹Mavg=38.32=x×44+(100−x)×28100 ⟹x=64.5%.